Question

A tennis ball is thrown horizontally from an elevation of \(14.0 \mathrm{m}\) above the ground with a speed of \(20.0 \mathrm{m} / \mathrm{s}\) (a) Where is the ball after 1.60 s? (b) If the ball is still in the air, how long before it hits the ground and where will it be with respect to the starting point once it lands?

Short answer

Answer: After 1.60 s, the tennis ball is 32.0 m horizontally away from the starting point and 5.664 m above the ground. It takes 1.69 s for the ball to hit the ground, and it will be 33.8 m away from the starting point when it lands.

Step-by-step solution

Identifying the knowns and unknowns

We are given the following information: - Initial elevation: \(h = 14.0 \mathrm{m}\) - Initial horizontal speed: \(v_0 = 20.0 \mathrm{m}/\mathrm{s}\) - Time: \(t = 1.60 \mathrm{s}\) We need to find the position of the ball after 1.60 s and the time it takes for the ball to hit the ground and its position at that moment.

Horizontal motion

For horizontal motion, there is no acceleration. Hence, the horizontal position (\(x\)) can be easily calculated using the formula: \(x = v_0 \times t\) Where \(v_0\) is the initial horizontal speed and t is the time.

Vertical motion

For vertical motion, the acceleration due to gravity, \(g = 9.8 \mathrm{m}/\mathrm{s}^2\), acts downward. The initial vertical speed is 0. We can find the vertical position (\(y\)) of the ball using the following equation: \(y = h + v_{0_y}t - \frac{1}{2}gt^2\) Where \(v_{0_y}\) is the initial vertical speed, which equals 0 in our case.

Calculate the position after 1.60 s

Now, we can find the position of the ball after 1.60 s using the formulas: For horizontal position: \(x = v_0 \times t = (20.0 \mathrm{m}/\mathrm{s}) \times (1.60 \mathrm{s}) = 32.0 \mathrm{m}\) For vertical position: \(y = h - \frac{1}{2}gt^2 = 14.0 \mathrm{m} - \frac{1}{2}(9.8 \mathrm{m}/\mathrm{s}^2)(1.60 \mathrm{s})^2 = 5.664 \mathrm{m}\) So, the ball is \(32.0 \mathrm{m}\) horizontally away from the starting point and \(5.664 \mathrm{m}\) above the ground after \(1.60 \mathrm{s}\).

Calculate the time it takes for the ball to hit the ground

To find when the ball hits the ground, we need to calculate when its vertical position (y) becomes 0: \(h - \frac{1}{2}gt^2 = 0\) Solve for time \(t\): \(t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(14.0 \mathrm{m})}{9.8\mathrm{m}/\mathrm{s}^2}} = 1.69 \mathrm{s}\)

Calculate the horizontal position when the ball hits the ground

Now that we know the time it takes for the ball to hit the ground, we can calculate the horizontal position at that moment: \(x = v_0 \times t = (20.0 \mathrm{m}/\mathrm{s}) \times (1.69 \mathrm{s}) = 33.8 \mathrm{m}\) So, the ball will hit the ground \(33.8 \mathrm{m}\) away from the starting point. To summarize, (a) The ball is 32.0 m horizontally away from the starting point and 5.664 m above the ground after 1.60 s. (b) The ball will take 1.69 s to hit the ground, and it will be 33.8 m away from the starting point when it lands.