Question

A tennis player tosses a tennis ball straight up and then catches it after 2.00 s at the same height as the point of release. (a) What is the acceleration of the ball while it is in flight? (b) What is the velocity of the ball when it reaches its maximum height? Find (c) the initial velocity of the ball and (d) the maximum height it reaches

Answer 1

Answer:

Part a)

a = -9.81 m/s/s

Part b)

v = 0

Part c)

v = 9.81 m/s

Part d)

$H = 4.905 m$

Explanation:

Part a)

During the motion of ball it will have only gravitational force on the ball

so here the acceleration of the ball is only due to gravity

so it is given as

$a = g = 9.81 m/s^2$

Part b)

As we know that ball is moving against the gravity

so here the velocity of ball will keep on decreasing as the ball moves upwards

so at the highest point of the motion of the ball the speed of ball reduce to zero

$v_f = 0$

Part c)

We know that the total time taken by the ball to come back to the initial position is T = 2 s

so in this time displacement of the ball will be zero

$\Delta y = 0 = v_y t + \frac{1}{2} at^2$

$0 = v_y (2) - \frac{1}{2}(9.81)(2^2)$

$v_y = 9.81 m/s$

Part d)

at the maximum height position we know that the final speed will be zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

here we have

$0 - (9.81^2) = 2(-9.81)H$

$H = 4.905 m$