Question

A ball is thrown vertically upward from the top of a tower with an initial velocity of 19.6 metre per second. The ball reaches the ground after 5s. Calculate

A. The height of the tower

B. The velocity of ball reaching the ground

Answer 1

Given,

Final velocity in upward journey, v = 0
Initial velocity, u = 19.6 m/s
total time in complete journey, t = 5 sec
$g=9.8m/s^2$

(A) Using the first equation of motion, v = u + at
t= u/g
= 19.6/9.8
=2 sec to go up
Therefore time to come down = 3 sec
Height reached by the ball above the tower = $h=u^2/2g$
$h=(19.6{)}^2/2\ast 9.8$
$h=19.6m$
Total height from the ground =$H=\left(1/2\right)g{t}^2=\left(1/2\right)\left(9.8\right)\ast (3{)}^2=44.1m$
Thus, the height of the building =44.1-19.6 = 24.5 m.

(B) Using the first equation of motion, v = u + at
v = gt
v = 9.8*3
v = 29.4 m/s.