A tennis ball is dropped from 1.99 m above the
ground. It rebounds to a height of 0.962 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/
. (Let
down be negative.)
Answer in units of m/s.
Can someone PLEASE HELP ME UNDERSTAND this.
Why did the tennis ball bring a ladder to the match?
Because it heard the competition was fierce and wanted to reach new heights!
Now, let's solve this problem together. The total change in height is given by the formula:
π = π’π + ππ’π
Where π is the total change in height, π’π is the initial velocity, and π’π is the final velocity. Since the ball bounces back, it reaches a height of 0.962 m, which is half of the total change in height.
So, π = 2 Γ 0.962 m = 1.924 m
Now, let's use the equation of motion to find the initial velocity:
(π’π)Β² = (π’π)Β² + 2ππ
Since the ball is freefalling, the acceleration is -9.8 m/sΒ² (negative due to it being downwards).
(π’π)Β² = (π’π)Β² + 2 Γ (-9.8 m/sΒ²) Γ 1.924 m
(π’π)Β² = (π’π)Β² - 37.9048 mΒ²/sΒ²
As the ball hits the ground, its final velocity is 0 m/s. Therefore, we have:
0 = (π’π)Β² - 37.9048 mΒ²/sΒ²
Rearranging this equation, we can solve for the initial velocity:
(π’π)Β² = 37.9048 mΒ²/sΒ²
π’π = β37.9048 m/s
π’π β 6.157 m/s
Hence, the velocity at which the tennis ball hits the ground is approximately 6.157 m/s.
Sure! I can help you understand this problem.
To solve this problem, we need to apply the principles of kinematics, specifically the equations of motion. In this case, we can use the equation for the height of an object in free fall:
h = v^2 / (2g)
where
h is the height,
v is the velocity, and
g is the acceleration due to gravity.
First, let's calculate the initial velocity of the ball when it touches the ground. We know that the ball is dropped from a height of 1.99 m, so the initial height (h1) is 1.99 m. We also know that the ball rebounds to a height of 0.962 m, so the final height (h2) is 0.962 m.
For the first part of the motion (ball falling), the initial height (h1) is positive, and the final height (h2) is 0. To solve for the initial velocity (v1), we can rearrange the equation:
v1 = sqrt(2gh1)
Using the given values, we have:
g = 9.8 m/s^2, and
h1 = 1.99 m.
Plugging in the values, we find:
v1 = sqrt(2 * 9.8 * 1.99) β 8.866 m/s.
For the second part of the motion (ball rebounding), the initial height (h1) is 0, and the final height (h2) is positive (0.962 m). To solve for the final velocity (v2), we can rearrange the equation:
v2 = sqrt(2gh2)
Using the given values, we have:
g = 9.8 m/s^2, and
h2 = 0.962 m.
Plugging in the values, we find:
v2 = sqrt(2 * 9.8 * 0.962) β 3.107 m/s.
Since the velocity is negative when the ball is coming back up, we need to take the negative value.
Therefore, the velocity with which the ball hits the ground (v3) is the negative final velocity (v2):
v3 = -v2 β -3.107 m/s.
So, the velocity with which the ball hits the ground is approximately -3.107 m/s.