Example Question #1 : Motion In Two Dimensions
An object is shot from the ground at 75m/s at an angle of 45⁰ above the horizontal. How high does the object get before beginning its descent?
The velocity must be broken down into x (horizontal) and y (vertical) components. We can use the y component to find how high the object gets. To find vertical velocity, vy, use .
Next we find how long it takes to reach the top of its trajectory using .
t = 5.3s
Finally, find how high the object goes with .
Example Question #283 : Mcat Physical Sciences
An object is shot from the ground at 125m/s at an angle of 30o above the horizontal. How far away does the object land?
First, find the horizontal (x) and vertical (y) components of the velocity
Next, find how long the object is in the air by calculating the time it takes it to reach the top of its path, and doubling that number.
t = 6.25s
Total time in the air is therefore 12.5s (twice this value).
Finally, find distance traveled my multiplying horizontal velocity and time.
Example Question #3 : Motion In Two Dimensions
A 2kg box is at the top of a frictionless ramp at an angle of 60o. The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.
What is the velocity of the box just before it hits the ground?
We can start this problem by determining how much time it takes the box the reach the ground. Since the vertical distance is 30m and we know the angle of the ramp.
Now that we have the distance traveled, we can determine the acceleration on the box due to gravity, using the equation .
We can plug these values into the following distance equation and solve for time.
Now that we know the acceleration on the box and the time of travel, we can use the equation to solve for the velocity.
Example Question #21 : Motion
A ball gets pushed off a high table with a horizontal speed of . How far does the ball travel horizontally before hitting the ground?
This is a two step problem. The first step is to calculate the time it takes for the ball to reach the ground. To find this time, we use the following kinematic equation dealing with vertical motion.
Choosing the ground to be the zero height, we have and .
Also, knowing that the initial vertical velocity is zero, we know that .
The kinematic equation simplifies using these values.
Rearrange the equation to isolate time.
We know that is the acceleration due to gravity: . Plug in the values to solve for time.
We now have the time the ball is travelling before it hits the ground. Use this value to find the horizontal distance before it hits the ground with the kinematic equation .
We know that and that . Using these values and the time, we can solve for the horizontal distance travelled.
Example Question #2 : Motion In Two Dimensions
A car drives north at for , then turns east and drives at for . What is the magnitude and direction of the average velocity for the trip?
First, determine how far the car travels in each direction:
Now that we have the directional displacements, we can find the total displacement by using the Pythagorean Theorem.
Find the average velocity by dividing the total displacement by the total time.
Velocity is a vector, meaning it has both magnitude and direction. Now that we have the magnitude, we can find the direction by using trigonometry.
Use the north and the east directional displacements to find the angle.
Our final answer will be:
Example Question #3 : Motion In Two Dimensions
A ball is launched at an angle of above the horizontal with an initial velocity of . At what time is its vertical velocity ?
Any projectile has a vertical velocity of zero at the peak of its flight. To solve this question, we need to find the time that it takes the ball to reach this height. The easiest way is to solve for the initial vertical velocity using trigonometry, and then use the appropriate kinematics equation to determine the time.
We know that the final vertical velocity will be zero. We can solve for the initial vertical velocity using the given angle and total velocity.
Using this in our kinematics formula, we solve for the time.
Keep in mind that this is only the vertical velocity. The total velocity at the peak is not zero, since the ball will still have horizontal velocity.
Example Question #22 : Motion
A 2kg box is at the top of a frictionless ramp at an angle of . The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.
When the box is released, how long will it take the box to reach the ground?
We can start this problem by determining how far the box will travel on the ramp before hitting the ground. Since the vertical distance is 30m and we know the angle of the ramp, we can determine the length of the hypotenuse using the equation .
Now that we have the distance traveled, we can determine the acceleration on the box due to gravity. Because the box is on a sloped surface, the box will not experience the full acceleration of gravity, but will instead be accelerated at a value of . Since the angle is 60o, the acceleration on the box is
Finally, we can plug these values into the following distance equation and solve for time.