Question

At serve, a tennis player aims to hit the ball horizontally, as shown in the figure.
(a) What minimum speed is required for the ball to clear the 0.90-m-high net about 15.0 m
from the server if the ball is “launched” from a height of 2.50 m?
(b) Where will the ball land if it just clears the net (and will it be “good” in the sense that it
lands within 7.0 m of the net)?
(d) How long will it be in the air?

Answer 1

(a) The minimum speed required for the ball to clear the net is 26.3 m/s.

(b) The horizontal distance of the ball when it clears the net is 11.3 m

(c) The total time spent in the air by the ball is 1 s.

Time of motion

The time of motion of the tennis ball is calculated by using the following kinematic equation as shown below;

$h = h_0 + v_0_yt - \frac{1}{2} gt^2\\\\0.9 = 2.5+ 0 - \frac{1}{2} (9.8) t^2\\\\0.9 = 2.5 - 4.9t^2\\\\4.9t^2 = 1.6\\\\t^2 = \frac{1.6}{4.9} \\\\t^2 = 0.327\\\\t = \sqrt{0.327} \\\\t = 0.57 \ s$

Minimum speed

The minimum speed of the ball is calculated as follows;

$v_x = \frac{X}{t} \\\\v_x = \frac{15}{0.57} \\\\v_x = 26.3 \ m/s$

Time of motion from top of the high net

The time of motion from top of the high net is calculated as follows;

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.9}{9.8} } \\\\t = 0.43 \ s$

Horizontal range = 0.43 x 26.3 = 11.3 m

Total time in air = 0.57 s + 0.43 s = 1.0 s

Learn more about horizontal velocity here: https://brainly.com/question/24949996