SAT Subject Test in Physics - Linear Motion

Near the surface of the earth, a projectile is fired from a canon at an angle of $\theta$ degrees above the horizon and an initial velocity of meters per second. Which of the following expressions gives the time it takes the projectile to reach its maximum height?

$t = v \sin \frac{\theta }{g}$

$t = v^{2} \sin \frac{\theta }{g}$

$t = v \cos \frac{\theta }{g}$

$t = -v^{2} \sin \frac{\theta }{g}$

$t = v \cos \frac{\theta }{2g}$

Explanation

At the maximum height of projectile motion, $v_{y}=0$ and because this takes place near the surface of the earth we know which we can plug into the equation:

$v_{y} = v_{0} + at$

$0 = v_{0} + gt$

We can then rearrange for

$t = \frac{v_{0}}{g}$

Next substitute this value into the equation $v_{y} = v \sin \theta$ to get the correct answer

$t = v \sin \frac{\theta }{g}$

$t = v^{2} \sin \frac{\theta }{g}$

$t = v \cos \frac{\theta }{g}$

$t = -v^{2} \sin \frac{\theta }{g}$

$t = v \cos \frac{\theta }{2g}$

Explanation

At the maximum height of projectile motion, $v_{y}=0$ and because this takes place near the surface of the earth we know which we can plug into the equation:

$v_{y} = v_{0} + at$

$0 = v_{0} + gt$

We can then rearrange for

$t = \frac{v_{0}}{g}$

Next substitute this value into the equation $v_{y} = v \sin \theta$ to get the correct answer

$t = v \sin \frac{\theta }{g}$

$t = v^{2} \sin \frac{\theta }{g}$

$t = v \cos \frac{\theta }{g}$

$t = -v^{2} \sin \frac{\theta }{g}$

$t = v \cos \frac{\theta }{2g}$

Explanation

At the maximum height of projectile motion, $v_{y}=0$ and because this takes place near the surface of the earth we know which we can plug into the equation:

$v_{y} = v_{0} + at$

$0 = v_{0} + gt$

We can then rearrange for

$t = \frac{v_{0}}{g}$

Next substitute this value into the equation $v_{y} = v \sin \theta$ to get the correct answer

$t = v \sin \frac{\theta }{g}$

If air resistance is negligible, 8 seconds after it is released, what would be the velocity of a stone dropped from a helicopter that has a horizontal velocity of 60 meters per second?

$100\frac{m}{s}$

$60\frac{m}{s}$

$80\frac{m}{s}$

$140\frac{m}{s}$

$160\frac{m}{s}$

Explanation

We are looking for total velocity, which in this case has both a horizontal and vertical component.

Because the helicopter is flying horizontally we know $v_{y0}=0$

We can assume that this takes place near the surface of the earth so

We can plug this into the equation:

$v_{yf}=v_{y0} +at$

$v_{yf}=0 +(10)(8)$

$v_{yf}=80$

Next we must find the horizontal velocity. Because there are no additional forces, the horizontal velocity is the same as the initial horizontal velocity of the helicopter, so:

$v_{x}=60$

Next we must use vector addition to add the horizontal and vertical components of velocity. Because this horizontal and vertical velocities are perpendicular the sum will be the hypotenuse of a right triangle:

$60^{2}+80^{2}=v^{2}$

$v=100\frac{m}{s}$

If air resistance is negligible, 8 seconds after it is released, what would be the velocity of a stone dropped from a helicopter that has a horizontal velocity of 60 meters per second?

$100\frac{m}{s}$

$60\frac{m}{s}$

$80\frac{m}{s}$

$140\frac{m}{s}$

$160\frac{m}{s}$

Explanation

We are looking for total velocity, which in this case has both a horizontal and vertical component.

Because the helicopter is flying horizontally we know $v_{y0}=0$

We can assume that this takes place near the surface of the earth so

We can plug this into the equation:

$v_{yf}=v_{y0} +at$

$v_{yf}=0 +(10)(8)$

$v_{yf}=80$

Next we must find the horizontal velocity. Because there are no additional forces, the horizontal velocity is the same as the initial horizontal velocity of the helicopter, so:

$v_{x}=60$

$60^{2}+80^{2}=v^{2}$

$v=100\frac{m}{s}$

If air resistance is negligible, 8 seconds after it is released, what would be the velocity of a stone dropped from a helicopter that has a horizontal velocity of 60 meters per second?

$100\frac{m}{s}$

$60\frac{m}{s}$

$80\frac{m}{s}$

$140\frac{m}{s}$

$160\frac{m}{s}$

Explanation

We are looking for total velocity, which in this case has both a horizontal and vertical component.

Because the helicopter is flying horizontally we know $v_{y0}=0$

We can assume that this takes place near the surface of the earth so

We can plug this into the equation:

$v_{yf}=v_{y0} +at$

$v_{yf}=0 +(10)(8)$

$v_{yf}=80$

Next we must find the horizontal velocity. Because there are no additional forces, the horizontal velocity is the same as the initial horizontal velocity of the helicopter, so:

$v_{x}=60$

$60^{2}+80^{2}=v^{2}$

$v=100\frac{m}{s}$

Which of these is not an example of Newtonian mechanics?

$F_G=G\frac{m_1m_2}{r^2}$

$\Delta U=\Delta PE+\Delta KE$

Explanation

Newtonian mechanics apply to all objects of substantial mass travelling at significantly slower than the speed of light.

Newton's law of universal gravitation, Newton's second law, momentum, and the equation for mechanical energy all fall under Newtonian mechanics.

The mass-energy equivalence suggests that mass can change as the speed of an object (such as an electron) approaches the speed of light. Newtonian mechanics assume that mass is constant, and do not apply to objects approaching the speed of light.

Which of these is not an example of Newtonian mechanics?

$F_G=G\frac{m_1m_2}{r^2}$

$\Delta U=\Delta PE+\Delta KE$

Explanation

Newtonian mechanics apply to all objects of substantial mass travelling at significantly slower than the speed of light.

Newton's law of universal gravitation, Newton's second law, momentum, and the equation for mechanical energy all fall under Newtonian mechanics.

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

At what angle to the horizontal will the rock impact the ground?

$41.99^{\circ}$

$44.3^{\circ}$

$21.1^{\circ}$

$33.5^{\circ}$

$17.1^{\circ}$

Explanation

The question gives the total initial velocity, but we will need to find the horizontal and vertical components.

To find the horizontal velocity we use the equation $v_{i}*\cos(\Theta)=v_{x}$ .

We can plug in the given values for the angle and initial velocity to solve.

$30\frac{m}{s}*\cos(35^{\circ})=v_{x}$

$30\frac{m}{s}*(0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

We can find the vertical velocity using the equation $v_{i}* \sin(\Theta)=v_{iy}$ .

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

The horizontal velocity will not change during flight because there are no forces in the horizontal direction. The vertical velocity, however, will be affected. We need to solve for the final vertical velocity, then combine the vertical and horizontal vectors to find the total final velocity.

We know that the rock is going to travel a net distance of , as that is the distance between where the rock's initial and final positions. We now know the displacement, initial velocity, and acceleration, which will allow us to solve for the final velocity.

$v_f^2=v_1^2+2a\Delta y$

$v_f^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-10m)$

$v_f^2=(296.18\frac{m^2}{s^2})+196\frac{m^2}{s^2}$

$v_f^2=492.18\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{492.18\frac{m^2}{s^2}}$

$v_f=22.19\frac{m}{s}$

Because the rock is traveling downward, our velocity will be negative: $v_f=-22.19\frac{m}{s}$ .

Now that we know our final velocities in both the horizontal and vertical directions, we can find the angle created between the two trajectories. The horizontal and vertical velocities can be compared using trigonometry.

$\tan(\Theta)=\frac{v_y}{v_x}$ ,

$\tan^{-1}(\frac{v_y}{v_x})=\Theta$

Plug in our values and solve for the angle.

$\tan^{-1}(\frac{22.19\frac{m}{s}}{24.57\frac{m}{s}})=\Theta$

$\tan^{-1}(0.90)=\Theta$

$41.99^{\circ}=\Theta$

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

At what angle to the horizontal will the rock impact the ground?

$41.99^{\circ}$

$44.3^{\circ}$

$21.1^{\circ}$

$33.5^{\circ}$

$17.1^{\circ}$

Explanation

The question gives the total initial velocity, but we will need to find the horizontal and vertical components.

To find the horizontal velocity we use the equation $v_{i}*\cos(\Theta)=v_{x}$ .

We can plug in the given values for the angle and initial velocity to solve.

$30\frac{m}{s}*\cos(35^{\circ})=v_{x}$

$30\frac{m}{s}*(0.82)=v_{x}$

$24.57\frac{m}{s}=v_{x}$

We can find the vertical velocity using the equation $v_{i}* \sin(\Theta)=v_{iy}$ .

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

$v_f^2=v_1^2+2a\Delta y$

$v_f^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-10m)$

$v_f^2=(296.18\frac{m^2}{s^2})+196\frac{m^2}{s^2}$

$v_f^2=492.18\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{492.18\frac{m^2}{s^2}}$

$v_f=22.19\frac{m}{s}$

Because the rock is traveling downward, our velocity will be negative: $v_f=-22.19\frac{m}{s}$ .

$\tan(\Theta)=\frac{v_y}{v_x}$ ,

$\tan^{-1}(\frac{v_y}{v_x})=\Theta$

Plug in our values and solve for the angle.

$\tan^{-1}(\frac{22.19\frac{m}{s}}{24.57\frac{m}{s}})=\Theta$

$\tan^{-1}(0.90)=\Theta$

$41.99^{\circ}=\Theta$

Linear Motion

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