Falling Objects with Drag - Differential Equations

1. A ball is dropped from rest in a medium where air resistance is proportional to velocity. The equation of motion is $\frac{dv}{dt} = g - kv$, where $g = 10 \text{ m/s}^2$ and $k = 0.5 \text{ s}^{-1}$. If the ball reaches 95% of its terminal velocity after time $t$, what is the approximate value of $t$?
   Answer: $t \approx 6.0 \text{ s}$ First find terminal velocity: $v_t = g/k = 10/0.5 = 20 \text{ m/s}$. The solution to the ODE is $v(t) = v_t(1 - e^{-kt}) = 20(1 - e^{-0.5t})$. For 95% of terminal velocity: $0.95 \times 20 = 20(1 - e^{-0.5t})$, so $0.95 = 1 - e^{-0.5t}$, giving $e^{-0.5t} = 0.05$. Taking natural log: $-0.5t = \ln(0.05) \approx -3.0$, so $t \approx 6.0 \text{ s}$. Choice A uses $\ln(0.05)$ directly without dividing by $k$. Choice C incorrectly uses $\ln(0.95)$. Choice D doubles the correct answer due to sign error.
2. The velocity of an object falling from rest with linear air resistance is given by the function $v(t) = A(1 - e^{-Bt})$, where $A$ and $B$ are positive constants. Which of the following correctly identifies the physical meaning of these constants in terms of the object's mass $m$, the acceleration due to gravity $g$, and the drag coefficient $k$?
   Answer: $A$ is the terminal velocity $mg/k$, and $B$ is the reciprocal of the time constant, $k/m$. The standard solution to the differential equation $m(dv/dt) = mg - kv$ with initial condition $v(0)=0$ is $v(t) = \frac{mg}{k}(1 - e^{-(k/m)t})$. Comparing this to the given form $v(t) = A(1 - e^{-Bt})$, we can identify the constants by matching the terms. The pre-factor $A$ corresponds to $\frac{mg}{k}$, which is the terminal velocity $v_T$. The constant $B$ in the exponent corresponds to $\frac{k}{m}$. The time constant is defined as $\tau = m/k$, so $B = 1/\tau$. Therefore, $A$ is the terminal velocity and $B$ is the ratio of the drag coefficient to the mass (the reciprocal of the time constant).
3. A skydiver of mass $m$ falls from rest. Air resistance is proportional to her velocity $v$, with proportionality constant $k$. At time $t_1$, her speed is $v_1$, and she opens her parachute, which instantly changes the drag coefficient to $k_2$, where $k_2 > k$. Which statement correctly describes her motion at the exact moment she opens the parachute?
   Answer: If her speed $v_1$ was greater than the new terminal velocity $mg/k_2$, she will decelerate. The differential equation for velocity is $m(dv/dt) = mg - kv$. At the moment the parachute opens, the drag coefficient $k$ changes to $k_2$. The velocity $v_1$ is continuous at this instant, but the acceleration changes to $dv/dt = g - (k_2/m)v_1$. The new terminal velocity is $v_{T2} = mg/k_2$. The skydiver will decelerate if $dv/dt < 0$, which means $g - (k_2/m)v_1 < 0$, or $mg < k_2v_1$. This is equivalent to $mg/k_2 < v_1$, or $v_{T2} < v_1$. Therefore, she decelerates only if her speed was greater than the new terminal velocity. Velocity cannot change instantaneously, so A is wrong. It's possible for $v_1$ to be less than $v_{T2}$, in which case she would continue to accelerate (but more slowly), so B is wrong. Since $k$ changes discontinuously, the acceleration $dv/dt = g - (k/m)v$ must also be discontinuous, so D is wrong.
4. The velocity of an object of mass $m$ dropped from rest, subject to linear drag with coefficient $k$, is given by $v(t) = \frac{mg}{k}(1 - e^{-kt/m})$. What is the limiting value of this expression as the drag coefficient $k$ approaches zero (i.e., the case of no drag)?
   Answer: $gt$ To find the limit of $v(t)$ as $k \to 0$, we can use L'Hôpital's Rule because the expression takes the indeterminate form $0/0$. Let's rewrite the expression as $v(t) = mg \cdot \frac{1 - e^{-kt/m}}{k}$. We evaluate the limit of the fraction: $\lim_{k\to 0} \frac{1 - e^{-kt/m}}{k}$. Differentiating the numerator and the denominator with respect to $k$, we get: Numerator derivative: $-(e^{-kt/m})(-t/m) = (t/m)e^{-kt/m}$. Denominator derivative: $1$. The limit becomes $\lim_{k\to 0} \frac{(t/m)e^{-kt/m}}{1} = (t/m)e^0 = t/m$. Therefore, the original limit is $mg \cdot (t/m) = gt$. This is the well-known formula for velocity under constant gravitational acceleration with no air resistance, which is the expected physical result.
5. An object is dropped from rest, and its velocity is described by $v(t) = v_T(1-e^{-t/\tau})$, where $v_T$ is the terminal velocity and $\tau$ is the time constant. After a very long time ($t \gg \tau$), the distance the object has fallen, $x(t)$, can be approximated by a linear function $x(t) \approx At - B$. What are the constants $A$ and $B$?
   Answer: $A = v_T$, $B = v_T \tau$ To find the distance $x(t)$, we integrate the velocity function $v(t)$ from $0$ to $t$, with $x(0)=0$. $x(t) = \int_0^t v_T(1-e^{-s/\tau}) ds = v_T [s + \tau e^{-s/\tau}]_0^t = v_T ((t + \tau e^{-t/\tau}) - (0 + \tau e^0)) = v_T(t - \tau + \tau e^{-t/\tau})$. For very long times, $t \gg \tau$, the exponential term $e^{-t/\tau}$ approaches zero. Therefore, the position function can be approximated as $x(t) \approx v_T(t - \tau) = v_T t - v_T \tau$. Comparing this to the form $At - B$, we find that $A = v_T$ and $B = v_T \tau$. This approximation represents the distance the object would have traveled if it had been moving at terminal velocity for a time $t$, minus a 'head start' distance of $v_T \tau$.
6. Two balls of different masses ($m_1 = 2 \text{ kg}$ and $m_2 = 8 \text{ kg}$) but identical shape and size are dropped simultaneously from the same height in air. Both experience linear drag with the same coefficient $k = 0.5 \text{ s}^{-1}$. After sufficient time, what is the ratio of their velocities $v_1/v_2$?
   Answer: $v_1/v_2 = 1:4$ because heavier objects fall faster in resistive media When you encounter problems involving objects falling through resistive media, the key insight is that terminal velocity depends on the balance between gravitational force and drag force. At terminal velocity, these forces are equal, so the object stops accelerating. For linear drag, the drag force is $F_d = kv$, where $k$ is the drag coefficient and $v$ is velocity. At terminal velocity, we have: $mg = kv_{terminal}$ Therefore: $v_{terminal} = \frac{mg}{k}$ Since both balls have the same drag coefficient $k = 0.5 \text{ s}^{-1}$, their terminal velocities are: - Ball 1: $v_1 = \frac{(2)(9.8)}{0.5} = 39.2 \text{ m/s}$ - Ball 2: $v_2 = \frac{(8)(9.8)}{0.5} = 156.8 \text{ m/s}$ The ratio is $v_1/v_2 = \frac{2g/k}{8g/k} = \frac{2}{8} = \frac{1}{4}$, confirming answer D. Let's examine why the other options are wrong: A is incorrect because terminal velocity clearly depends on mass in the formula $v_{terminal} = \frac{mg}{k}$. B reverses the relationship—while lighter objects do experience relatively more drag per unit mass, this actually makes them fall slower, not faster. C suggests a square root relationship, but linear drag produces a direct proportionality between terminal velocity and mass. Study tip: Remember that terminal velocity always equals $\frac{mg}{\text{drag coefficient}}$. For linear drag, velocity is directly proportional to mass; for quadratic drag ($F_d = kv^2$), it's proportional to the square root of mass.
7. A particle of mass $m$ is projected downward into a fluid with an initial speed of $10$ m/s. The net downward force on the particle is given by $F_{net} = mg - B - kv$, where $B$ is a constant buoyant force and $k$ is a positive drag coefficient. The particle's terminal velocity is observed to be $v_T = 5$ m/s. Which of the following describes the subsequent motion of the particle?
   Answer: The particle will slow down, with its speed approaching $5$ m/s from above. The equation of motion is $m(dv/dt) = (mg-B) - kv$. The terminal velocity $v_T$ is the speed at which the net force is zero, so $mg - B - kv_T = 0$. This implies that the term $(mg-B)$ is equal to $kv_T$. Substituting this back into the equation of motion gives $m(dv/dt) = kv_T - kv = k(v_T - v)$. The initial velocity is $v(0) = 10$ m/s, and the terminal velocity is $v_T = 5$ m/s. Since $v(0) > v_T$, the term $(v_T - v)$ is initially negative. This means $dv/dt$ is negative, so the particle decelerates. The velocity will decrease from $10$ m/s, asymptotically approaching the stable equilibrium velocity $v_T = 5$ m/s.
8. A particle falls from rest through a medium with linear drag. The velocity as a function of time is $v(t) = 20(1 - e^{-0.3t}) \text{ m/s}$. At what time does the particle's acceleration equal $3 \text{ m/s}^2$?
   Answer: $t = \frac{1}{0.3}\ln(2) \approx 2.31 \text{ s}$ from direct acceleration analysis When you encounter a particle motion problem with a given velocity function, remember that acceleration is simply the derivative of velocity with respect to time. This connects kinematics directly to calculus. Given $v(t) = 20(1 - e^{-0.3t})$, you find acceleration by taking the derivative: $a(t) = \frac{dv}{dt} = 20 \cdot \frac{d}{dt}(1 - e^{-0.3t}) = 20 \cdot (0 - (-0.3)e^{-0.3t}) = 6e^{-0.3t}$ To find when $a(t) = 3 \text{ m/s}^2$, set up the equation: $6e^{-0.3t} = 3$ Solving: $e^{-0.3t} = \frac{1}{2}$ Taking the natural logarithm: $-0.3t = \ln(\frac{1}{2}) = -\ln(2)$ Therefore: $t = \frac{\ln(2)}{0.3} \approx 2.31 \text{ s}$ This confirms answer C is correct. A uses $\ln(1.5)$, which would arise from incorrectly setting $6e^{-0.3t} = 4$ instead of 3. B uses $\ln(3)$, which comes from mistakenly setting $e^{-0.3t} = \frac{1}{3}$ rather than $\frac{1}{2}$. D uses $\ln(4)$, suggesting an error where someone set $e^{-0.3t} = \frac{1}{4}$. Study tip: Always remember that acceleration equals $\frac{dv}{dt}$. When working with exponential functions, double-check your algebra when isolating the exponential term—small errors in fractions lead to completely different logarithmic answers.
9. An object of mass $m=5$ kg is falling with a speed of $v_0 = 10$ m/s at time $t=0$. It is subject to gravity ($g=10$ m/s²) and a linear drag force $F_d = -2.5v$. What is the speed of the object at $t=2$ seconds?
   Answer: $20 - 10e^{-1}$ m/s The equation of motion is $m(dv/dt) = mg - kv$. Plugging in the values, we have $5(dv/dt) = 5(10) - 2.5v$, which simplifies to $dv/dt = 10 - 0.5v$. The terminal velocity is $v_T = mg/k = 50/2.5 = 20$ m/s. The general solution is of the form $v(t) = v_T + C e^{-(k/m)t}$. Here, $k/m = 2.5/5 = 0.5$, so $v(t) = 20 + C e^{-0.5t}$. We use the initial condition $v(0)=10$ to find $C$: $10 = 20 + C e^0$, which gives $C = -10$. So, the specific solution is $v(t) = 20 - 10e^{-0.5t}$. To find the speed at $t=2$, we evaluate $v(2) = 20 - 10e^{-0.5(2)} = 20 - 10e^{-1}$. Choice A comes from incorrectly using $v(0)=0$. Choice C comes from incorrectly using $m/k$ instead of $k/m$ in the exponent.
10. A falling object with linear drag has the differential equation $\frac{dv}{dt} + kv = g$. If $k = 0.4 \text{ s}^{-1}$ and $g = 10 \text{ m/s}^2$, and the object is dropped from rest, what is the acceleration when the velocity reaches half the terminal velocity?
    Answer: $a = 5.0 \text{ m/s}^2$ because half-velocity creates half-maximum drag When you encounter a differential equation for motion with drag, remember that acceleration equals the net force divided by mass, which translates directly to the right side of your DE when written in standard form. The given equation $\frac{dv}{dt} + kv = g$ can be rewritten as $\frac{dv}{dt} = g - kv$. Since $\frac{dv}{dt}$ is acceleration, we have $a = g - kv$. First, find the terminal velocity where acceleration becomes zero: $0 = g - kv_t$, so $v_t = \frac{g}{k} = \frac{10}{0.4} = 25 \text{ m/s}$. When velocity reaches half the terminal velocity: $v = \frac{25}{2} = 12.5 \text{ m/s}$ Substituting into the acceleration equation: $a = 10 - 0.4(12.5) = 10 - 5 = 5.0 \text{ m/s}^2$ Answer D correctly recognizes that at half terminal velocity, the drag force is exactly half its maximum value, leaving half the gravitational acceleration unbalanced. Answer A uses the wrong fraction—at half velocity, drag reduces acceleration by half of its maximum effect, not to one-quarter of gravitational acceleration. Answer B ignores that drag increases with velocity, incorrectly assuming constant gravitational acceleration throughout the fall. Answer C miscalculates by not properly accounting for the linear relationship between velocity and drag force. Key strategy: For linear drag problems, always remember that $a = g - kv$. Terminal velocity occurs when these balance, so at any fraction of terminal velocity, you can quickly calculate the remaining unbalanced acceleration.
11. Two spherical objects, Sphere 1 and Sphere 2, are made of the same uniform material. Sphere 1 has twice the radius of Sphere 2 ($r_1 = 2r_2$). They are dropped from rest in the same fluid, where the drag force is linear with velocity, $F_d = -kv$, and the drag coefficient $k$ is directly proportional to the radius $r$ (i.e., $k=cr$ for some constant $c$). How do their terminal velocities, $v_{T1}$ and $v_{T2}$, compare?
    Answer: $v_{T1} = 4v_{T2}$ The terminal velocity is given by the formula $v_T = mg/k$. The mass $m$ of a sphere is its density $\rho$ times its volume $V = \frac{4}{3}\pi r^3$. Since the material is the same, $\rho$ is constant, so $m$ is proportional to $r^3$. Thus, $m_1/m_2 = (r_1/r_2)^3 = 2^3 = 8$. We are given that the drag coefficient $k$ is proportional to the radius $r$. Thus, $k_1/k_2 = r_1/r_2 = 2$. Now we can find the ratio of the terminal velocities: $\frac{v_{T1}}{v_{T2}} = \frac{m_1 g / k_1}{m_2 g / k_2} = \frac{m_1}{m_2} \cdot \frac{k_2}{k_1}$. Substituting the ratios we found: $\frac{v_{T1}}{v_{T2}} = (8) \cdot \frac{1}{2} = 4$. Therefore, $v_{T1} = 4v_{T2}$.
12. An object is thrown vertically upward with an initial speed $v_0$. It is subject to both gravity and linear air resistance. Which statement is true about its motion?
    Answer: The time taken to reach the maximum height is less than the time it would take in a vacuum. Let downward be the positive direction. The equation of motion is $m(dv/dt) = mg - kv$. On the way up, velocity $v$ is negative, so the drag force $-kv$ is positive (downward). Thus, drag and gravity both act downward, so the magnitude of acceleration is $|a| = |g - (k/m)v| = g + (k/m)|v| > g$. In a vacuum, the acceleration magnitude is always $g$. Since the upward-moving object with drag has a greater downward acceleration, it will slow down more quickly and reach its maximum height in less time than in a vacuum. A and B are false because air resistance is a non-conservative force that removes mechanical energy from the system. C is false because the acceleration magnitude $g + (k/m)|v|$ is greatest when $|v|$ is greatest, which occurs at the start of the motion, not at the peak where $v=0$.
13. A small sphere is dropped from rest in a viscous fluid. It is observed to reach 50% of its terminal velocity in $\ln(2)$ seconds. Assuming the drag force is linear with velocity ($F_d = -kv$), what is the time constant $\tau = m/k$ for this system?
    Answer: $1$ s The velocity of an object falling from rest with linear drag is given by the equation $v(t) = v_T(1 - e^{-t/\tau})$, where $v_T$ is the terminal velocity and $\tau$ is the time constant. We are given that at $t = \ln(2)$, the velocity is $v(t) = 0.5 v_T$. Substituting these values into the equation gives: $0.5 v_T = v_T(1 - e^{-\ln(2)/\tau})$. Dividing by $v_T$, we get $0.5 = 1 - e^{-\ln(2)/\tau}$. Rearranging yields $e^{-\ln(2)/\tau} = 0.5$. Taking the natural logarithm of both sides, we have $-\ln(2)/\tau = \ln(0.5) = -\ln(2)$. Therefore, $\ln(2)/\tau = \ln(2)$, which implies that $\tau = 1$ second.
14. An object of mass $m$ and drag coefficient $k$ is dropped from rest in a fluid. Let $v_T = mg/k$ be the terminal velocity and $\tau = m/k$ be the time constant of the system. What is the total distance the object falls in a time interval equal to two time constants, i.e., $t = 2\tau$?
    Answer: $v_T \tau (1 + e^{-2})$ First, find the velocity function: $v(t) = v_T(1 - e^{-t/\tau})$. To find the distance fallen, integrate the velocity from $t=0$ to $t=2\tau$. Let $x(t)$ be the distance. $x(t) = \int_0^t v(s) ds = \int_0^t v_T(1 - e^{-s/\tau}) ds = v_T [s + \tau e^{-s/\tau}]_0^t = v_T((t + \tau e^{-t/\tau}) - (0 + \tau e^0)) = v_T(t + \tau e^{-t/\tau} - \tau)$. Now, substitute $t = 2\tau$: $x(2\tau) = v_T(2\tau + \tau e^{-2\tau/\tau} - \tau) = v_T(\tau + \tau e^{-2}) = v_T \tau (1 + e^{-2})$. Distractor B results from an integration constant error. Distractor C represents the velocity at $t=2\tau$, not the distance. Distractor D is the distance that would be covered if the object traveled at its terminal velocity for the entire duration.
15. A sphere falls through a viscous fluid where the drag force follows $F_d = kv$. The equation of motion is $m\frac{dv}{dt} = mg - kv$. If the mass is doubled while keeping the drag coefficient constant, how does the time to reach $95\%$ of terminal velocity change?
    Answer: The time doubles because the time constant $\tau = m/k$ doubles proportionally The time constant is $\tau = m/k$. The approach to terminal velocity follows $v(t) = v_t(1 - e^{-t/\tau})$. To reach $95\%$: $0.95 = 1 - e^{-t/\tau}$, so $t = \tau\ln(20)$. When mass doubles, $\tau$ doubles, so the time to reach $95\%$ also doubles. The terminal velocity also doubles ($v_t = mg/k$), but this doesn't affect the time scale. Choice B confuses mass with acceleration effects. Choice C ignores the time constant change. Choice D applies incorrect scaling.
16. A skydiver experiences two phases of motion: free fall with minimal drag ($k_1 \approx 0$) followed by descent with significant linear drag ($k_2 = 0.8 \text{ s}^{-1}$) after reaching velocity $v_1 = 60 \text{ m/s}$. In the second phase, what is the velocity after $2.5$ seconds if the terminal velocity is $50 \text{ m/s}$?
    Answer: $v = 50 + 10e^{-2.0} \approx 51.35 \text{ m/s}$ using exponential approach from above When analyzing motion with air resistance, you need to set up a differential equation that accounts for how drag force affects acceleration. For a falling object with linear drag, the equation is $\frac{dv}{dt} = g - kv$, where terminal velocity occurs when $g = kv_t$. Since terminal velocity is 50 m/s and $k_2 = 0.8 \text{ s}^{-1}$, we can verify: $g = 0.8 \times 50 = 40 \text{ m/s}^2$. The differential equation becomes $\frac{dv}{dt} = 40 - 0.8v$. Rearranging: $\frac{dv}{dt} = -0.8(v - 50)$. This separable equation gives us $v(t) = 50 + (v_0 - 50)e^{-0.8t}$. Since the skydiver enters the second phase at $v_0 = 60 \text{ m/s}$, we have $v(t) = 50 + 10e^{-0.8t}$. After 2.5 seconds: $v = 50 + 10e^{-2.0} \approx 51.35 \text{ m/s}$. This is answer C. Answer A incorrectly models simple exponential decay without considering terminal velocity equilibrium. Answer B assumes the skydiver starts below terminal velocity and approaches from below, contradicting the given initial condition of 60 m/s. Answer D uses the wrong equilibrium point, treating the initial velocity as the target rather than terminal velocity. The key insight is recognizing that when an object starts above terminal velocity, it exponentially approaches that terminal velocity from above. Always identify whether the initial condition is above or below the equilibrium value to determine the direction of approach.
17. The velocity of a falling object with linear drag is $v(t) = v_T(1 - e^{-t/\tau})$, where $v_T$ is terminal velocity and $\tau=m/k$ is the time constant. Which of the following statements provides the best physical interpretation of the time constant $\tau$?
    Answer: If the object maintained its initial acceleration, it would reach its terminal velocity $v_T$ in time $\tau$. The governing differential equation is $dv/dt = g - (k/m)v$. The initial acceleration (at $t=0$, when $v=0$) is $a(0) = g$. If the object were to continue accelerating at this constant rate, its velocity would be $v(t) = gt$. To reach terminal velocity, $v_T = mg/k$, would require time $t$ such that $mg/k = gt$. Solving for $t$ gives $t = m/k$, which is defined as the time constant $\tau$. So, $\tau$ is the time it would take to reach $v_T$ with constant initial acceleration $g$. Choice A is incorrect; that would be roughly $t=5\tau$. Choice B is incorrect; half of terminal velocity is reached at $t = \tau \ln(2)$. Choice D is incorrect; acceleration is $a(t)=ge^{-t/\tau}$, so it is halved when $e^{-t/\tau}=0.5$, which is at $t=\tau \ln(2)$
18. Consider the velocity-time relationship for a falling object with linear drag: $v(t) = v_t(1 - e^{-t/\tau})$ where $\tau = 1/k$ is the time constant. If the object reaches $63.2\%$ of its terminal velocity at time $t_1$ and $86.5\%$ at time $t_2$, what is the relationship between $t_1$ and $t_2$?
    Answer: $t_2 = 2t_1$ because the exponential relationship doubles the time interval At $63.2\% = 1 - 1/e \approx 0.632$: $0.632v_t = v_t(1 - e^{-t_1/\tau})$, so $e^{-t_1/\tau} = 0.368 = 1/e$, giving $t_1 = \tau$. At $86.5\% = 1 - 1/e^2 \approx 0.865$: $0.865v_t = v_t(1 - e^{-t_2/\tau})$, so $e^{-t_2/\tau} = 0.135 = 1/e^2$, giving $t_2 = 2\tau = 2t_1$. Choice B assumes linear relationship. Choice C incorrectly adds time constants. Choice D assumes quadratic approach.
19. Two identical spheres are dropped simultaneously from the same height, one in air ($k_1 = 0.2 \text{ s}^{-1}$) and one in water ($k_2 = 0.8 \text{ s}^{-1}$). Both experience linear drag proportional to velocity. After a long time, what is the ratio of their terminal velocities $v_{t,air}/v_{t,water}$?
    Answer: $4:1$ (the sphere in air has 4 times the terminal velocity) Terminal velocity occurs when $mg = kv_t$, so $v_t = mg/k$. Since the spheres are identical, they have the same mass and experience the same gravitational force. Therefore $v_{t,air}/v_{t,water} = k_2/k_1 = 0.8/0.2 = 4$. The sphere in air has 4 times the terminal velocity because it experiences less drag resistance. Choice A inverts the ratio. Choice C ignores the different drag coefficients. Choice D uses the square root of the correct ratio.

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