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020 (Part 1 of 3)

Problem: A tennis ball is dropped from a height of 1.97 m. It rebounds to a height of 0.811 m.
With what velocity does it hit the ground?
$g = 9.8 \, \text{m/s}^2$ (Let down be negative).

Solution:

We use the kinematic equation: $v_f^2 = v_i^2 + 2a d$ Here:

• $v_i = 0 \, \text{m/s}$ (initial velocity, since the ball is dropped).
• $a = -9.8 \, \text{m/s}^2$ (acceleration due to gravity, negative as downward).
• $d = -1.97 \, \text{m}$ (displacement, negative as it's downward).

We want to find the velocity with which the ball hits the ground, $v_f$.
Substitute the known values: $v_f^2 = 0 + 2(-9.8)(-1.97)$ $v_f^2 = 38.524$ $v_f = \sqrt{38.524} = 6.21 \, \text{m/s}$ So, the velocity with which the ball hits the ground is 6.21 m/s downward.

021 (Part 2 of 3)

Problem: With what velocity does it leave the ground?

Solution:

When the ball rebounds, the maximum height it reaches is $0.811 \, \text{m}$. We'll use the same kinematic equation to find the velocity with which the ball leaves the ground, but now upwards.

The equation is: $v_f^2 = v_i^2 + 2a d$ Here:

• $v_f = 0 \, \text{m/s}$ (final velocity at the top of the rebound, since the ball momentarily stops).
• $a = -9.8 \, \text{m/s}^2$ (gravity is still downward).
• $d = 0.811 \, \text{m}$ (positive because upward displacement).

Now we solve for $v_i$, the initial velocity: $0 = v_i^2 + 2(-9.8)(0.811)$ $v_i^2 = 2(9.8)(0.811) = 15.896$ $v_i = \sqrt{15.896} = 3.99 \, \text{m/s}$ So, the velocity with which the ball leaves the ground is 3.99 m/s upward.

022 (Part 3 of 3)

Problem: If the tennis ball were in contact with the ground for $0.00715 \, \text{s}$, find the acceleration given to the ball by the ground.

Solution:

We have the following information:

• Initial velocity when the ball hits the ground: $v_i = -6.21 \, \text{m/s}$ (downward).
• Final velocity when the ball leaves the ground: $v_f = 3.99 \, \text{m/s}$ (upward).
• Time of contact with the ground: $t = 0.00715 \, \text{s}$.

We can use the formula for acceleration: $a = \frac{\Delta v}{\Delta t}$ $a = \frac{v_f - v_i}{t} = \frac{3.99 - (-6.21)}{0.00715}$ $a = \frac{3.99 + 6.21}{0.00715} = \frac{10.2}{0.00715} = 1,426.57 \, \text{m/s}^2$ So, the acceleration given to the tennis ball by the ground is 1,426.57 m/s².

Would you like more details on any of these solutions? Here are some related questions for further practice:

1. How would the time of contact change if the ball rebounded to a lower height?
2. What happens to the acceleration if the ball was in contact with the ground for twice as long?
3. Can you calculate the impulse exerted on the ball during the collision with the ground?
4. How would you calculate the potential energy of the ball at its maximum height?
5. How would air resistance affect the ball's motion in this scenario?

Tip: Always double-check your signs when dealing with direction in kinematic problems!