Question
(part 1 of 3)
A tennis ball is dropped from 1.83 m above the
ground. It rebounds to a height of 0.974 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s^2. (Let
down be negative.)
Answer in units of m/s.
(part 2 of 3)
With what velocity does it leave the ground?
Answer in units of m/s.
(part 3 of 3)
If the tennis ball were in contact with the
ground for 0.0103 s, find the acceleration
given to the tennis ball by the ground.
A tennis ball is dropped from 1.83 m above the
ground. It rebounds to a height of 0.974 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s^2. (Let
down be negative.)
Answer in units of m/s.
(part 2 of 3)
With what velocity does it leave the ground?
Answer in units of m/s.
(part 3 of 3)
If the tennis ball were in contact with the
ground for 0.0103 s, find the acceleration
given to the tennis ball by the ground.
Answers
Answered by
GPT-4o mini
Let's solve each part step by step.
### Part 1: Velocity when hitting the ground
To find the velocity just before the ball hits the ground, we can use the following kinematic equation:
\[
v^2 = u^2 + 2as
\]
Where:
- \( v \) = final velocity (m/s)
- \( u \) = initial velocity (m/s) = 0 (since the ball is dropped)
- \( a \) = acceleration (m/s²) = -9.8 m/s² (downward)
- \( s \) = distance fallen (m) = -1.83 m (since down is negative)
Plugging in the values:
\[
v^2 = 0 + 2 \times (-9.8) \times (-1.83)
\]
Calculating \( 2 \times 9.8 \times 1.83 \):
\[
v^2 = 2 \times 9.8 \times 1.83 \approx 35.8808
\]
Now take the square root to find \( v \):
\[
v = \sqrt{35.8808} \approx 5.99 \, \text{m/s}
\]
Since down is negative, the velocity just before it hits the ground is:
\[
v \approx -5.99 \, \text{m/s}
\]
### Part 2: Velocity leaving the ground
To find the velocity just after the ball leaves the ground after rebounding to a height of 0.974 m, we again use the kinematic equation, but now the ball is moving upwards.
Using:
\[
v^2 = u^2 + 2as
\]
Where:
- \( v \) = final velocity (m/s)
- \( u \) = initial velocity (m/s), which we want to find
- \( a \) = acceleration = -9.8 m/s² (upwards after the rebound)
- \( s \) = height reached (m) = 0.974 m
Rearranging gives us:
\[
0 = u^2 - 2 \times 9.8 \times 0.974
\]
Calculating \( 2 \times 9.8 \times 0.974 \):
\[
u^2 = 2 \times 9.8 \times 0.974 \approx 19.0936
\]
Now take the square root to find \( u \):
\[
u = \sqrt{19.0936} \approx 4.37 \, \text{m/s}
\]
Since it is moving upwards, the velocity just after it leaves the ground is:
\[
u \approx 4.37 \, \text{m/s}
\]
### Part 3: Acceleration given to the tennis ball by the ground
To find the acceleration, we can use the formula:
\[
a = \frac{\Delta v}{\Delta t}
\]
Where:
- \( \Delta v \) = change in velocity = final velocity - initial velocity
- \( \Delta t \) = time in contact with the ground = 0.0103 s
The change in velocity \( \Delta v \) when hitting the ground (down is negative) to rebounding upward is:
\[
\Delta v = v_{\text{up}} - v_{\text{down}} = 4.37 - (-5.99) = 4.37 + 5.99 = 10.36 \, \text{m/s}
\]
Now plug this into the formula for acceleration:
\[
a = \frac{10.36}{0.0103} \approx 1003.88 \, \text{m/s}^2
\]
Thus, the acceleration given to the tennis ball by the ground is:
\[
a \approx 1003.88 \, \text{m/s}^2
\]
### Summary of Answers
1. Velocity upon hitting the ground: **-5.99 m/s**
2. Velocity after leaving the ground: **4.37 m/s**
3. Acceleration given by the ground: **1003.88 m/s²**
### Part 1: Velocity when hitting the ground
To find the velocity just before the ball hits the ground, we can use the following kinematic equation:
\[
v^2 = u^2 + 2as
\]
Where:
- \( v \) = final velocity (m/s)
- \( u \) = initial velocity (m/s) = 0 (since the ball is dropped)
- \( a \) = acceleration (m/s²) = -9.8 m/s² (downward)
- \( s \) = distance fallen (m) = -1.83 m (since down is negative)
Plugging in the values:
\[
v^2 = 0 + 2 \times (-9.8) \times (-1.83)
\]
Calculating \( 2 \times 9.8 \times 1.83 \):
\[
v^2 = 2 \times 9.8 \times 1.83 \approx 35.8808
\]
Now take the square root to find \( v \):
\[
v = \sqrt{35.8808} \approx 5.99 \, \text{m/s}
\]
Since down is negative, the velocity just before it hits the ground is:
\[
v \approx -5.99 \, \text{m/s}
\]
### Part 2: Velocity leaving the ground
To find the velocity just after the ball leaves the ground after rebounding to a height of 0.974 m, we again use the kinematic equation, but now the ball is moving upwards.
Using:
\[
v^2 = u^2 + 2as
\]
Where:
- \( v \) = final velocity (m/s)
- \( u \) = initial velocity (m/s), which we want to find
- \( a \) = acceleration = -9.8 m/s² (upwards after the rebound)
- \( s \) = height reached (m) = 0.974 m
Rearranging gives us:
\[
0 = u^2 - 2 \times 9.8 \times 0.974
\]
Calculating \( 2 \times 9.8 \times 0.974 \):
\[
u^2 = 2 \times 9.8 \times 0.974 \approx 19.0936
\]
Now take the square root to find \( u \):
\[
u = \sqrt{19.0936} \approx 4.37 \, \text{m/s}
\]
Since it is moving upwards, the velocity just after it leaves the ground is:
\[
u \approx 4.37 \, \text{m/s}
\]
### Part 3: Acceleration given to the tennis ball by the ground
To find the acceleration, we can use the formula:
\[
a = \frac{\Delta v}{\Delta t}
\]
Where:
- \( \Delta v \) = change in velocity = final velocity - initial velocity
- \( \Delta t \) = time in contact with the ground = 0.0103 s
The change in velocity \( \Delta v \) when hitting the ground (down is negative) to rebounding upward is:
\[
\Delta v = v_{\text{up}} - v_{\text{down}} = 4.37 - (-5.99) = 4.37 + 5.99 = 10.36 \, \text{m/s}
\]
Now plug this into the formula for acceleration:
\[
a = \frac{10.36}{0.0103} \approx 1003.88 \, \text{m/s}^2
\]
Thus, the acceleration given to the tennis ball by the ground is:
\[
a \approx 1003.88 \, \text{m/s}^2
\]
### Summary of Answers
1. Velocity upon hitting the ground: **-5.99 m/s**
2. Velocity after leaving the ground: **4.37 m/s**
3. Acceleration given by the ground: **1003.88 m/s²**