Question

A boy standing on the ground close to a building throws a ball vertically upward. From his measurements of the maximum height, y-max, to which the ball rises and the time required to reach this height, the boy calculates that the average velocity of the ball on its way up is 20 m/s. Five seconds after leaving the boys hand, the ball is caught by a girl who has stretched her arm out of a window some distance above the boy. You may use -9.8 m/s^2 for the value of g.

a. What is the velocity of the ball immediately after its release?

b. How high does the ball rise; i.e., what is the value of y-max?

c. At what height, y, above the point of release is the ball caught?

d. What is the velocity of the ball immediately before being caught

Answer 1

Average velocity of ball on its way up:  Vav= 20 m/s

:gravity  g= 9.8 m/s2

The initial velocity of ball Vi

The velocity of the ball at the highest point   Vf=0 m/s

Maximum height  H

Time taken by ball   T

Time taken in downward journey  t

Distance from top when the ball is caught:   s

Velocity immediately before the ball is caught  vf

(a)

Average velocity is given by:

Vav =(Vi+Vf)/2  =Vi+02  = Vi/2.

Putting the values

Vi=2×Vav  =2×20  =  40 m/s.

(b)

Let H be the y-max

H=Vi2/2g  =40.02/(2x9.8) =81.633 m

(c)

Now

T=Vi/g  =409.8  =4.082 s

The ball is caught 55 seconds after it leaves i.e.

t=5.0−4.082  =0.918 s

Distance traveled:

s=gt2/2  =9.8×0.91822  =4.13 m

Therefore the height at which the ball is caught is

h=81.63−4.13  =77.50 m

(d)

The velocity of the ball immediately before being caught is given by

vf=√(2gxs) =√2×9.8×4.13  ≈9.0 m/s

I hope this will help you. Have a good day buddy. Thanks.