Question

A tennis player tosses a tennis ball straight up and then catches it after 1.77 s at the same height as the point of release. (a) What is the acceleration of the ball while it is in flight? magnitude m/s2 direction (b) What is the velocity of the ball when it reaches its maximum height? magnitude m/s direction (c) Find the initial velocity of the ball. m/s upward (d) Find the maximum height it reaches. m'

Answer 1

(a) 9.8 m/s^2, downward

There is only one force acting on the ball while it is in flight: the force of gravity, which is

F = mg

where

m is the mass of the ball

g is the gravitational acceleration

According to Newton's second law, the force acting on the ball is equal to the product between the mass of the ball and its acceleration, so

F = mg = ma

which means

a = g

So, the acceleration of the ball during the whole flight is equal to the acceleration of gravity:

g = -9.8 m/s^2

where the negative sign means the direction is downward.

(b) v = 0

Any object thrown upward reaches its maximum height when its velocity is zero:

v = 0

In fact, at that moment, the object's velocity is turning from upward to downward: that means that at that instant, the velocity must be zero.

(c) 8.72 m/s, upward

The initial velocity of the ball can be found by using the equation:

v = u + at

Where

v = 0 is the velocity at the maximum height

u is the initial velocity

a = g = -9.8 m/s^2 is the acceleration

t is the time at which the ball reaches the maximum height: this is half of the time it takes for the ball to reach again the starting point of the motion, so

$t=\frac{1.77 s}{2}=0.89 s$

So we can now solve the equation for u, and we find:

$u=v-at=0-(-9.8 m/s^2)(0.89 s)=8.72 m/s$

(d) 3.88 m

The maximum height reached by the ball can be found by using the equation:

$v^2 - u^2 = 2ad$

where

v = 0 is the velocity at the maximum height

u = 8.72 m/s is the initial velocity

a = g = -9.8 m/s^2 is the gravitational acceleration

d is the maximum height reached

Solving the equation for d, we find

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(8.72 m/s)^2}{2(-9.8 m/s^2)}=3.88 m$