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
There are no human answers yet.
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**
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