Question

A ball is thrown vertically upwards from the top of a tower with an initial velocity of 19.6m/s. The ball reaches the ground after 5s. Calculate

1. The height of the tower

2. The velocity of ball on reaching the ground.

(Take g=9.8m/s)

Answer 1

Velocity of Ball At the topmost point is zero.
v=u+at.
[Ball thrown upwards, upward velocity is taken as +ve,so 'g' which acting in downward direction is taken as -ve.,a=-g,]
=>V=0
u=19.6m/s
total time=5sec
g=9.8m/s^2
t= u/g
= 19.6/9.8
=2sec to go up
Total time to rrach the ground=5s. Therefore time to come down =3sec .
Using,
(V^2)-(U^2)= 2as
at the topmost point, V=0,a=-g. Then,
From the top of the building to topmost point,
height = (u^2)/2g
​​​​​​=[(19.6)^2] / 2*9.8
=19.6m

• from the topmost point to the ground,
S=ut+1/2×at^2
Starting from Top most point.so u=0 ,a=g
(there is no velocity, so only the direction of g is there.so this is taken as +ve.]
Total height =1/2×g×t^2
=1/2× (9.8)×(3)^2
=44.1m
• height of the building =44.1-19.6
=24.5m

• v=gt
(considering initial velocity as the velocity at the topmost point which is zero,u=0. Time to reach the ground,t=3s]
= 9.8*3
=29.4m/s