An object is thrown vertically upwards with a velocity 30 m·s-1 from the top of a tower. After 4 s, from the same place, another object is dropped. If both of them touch the ground simultaneously, find (i) height of the tower and(ii) total time of fall of the second object. [g = 10 m.s-2]

Answers (1)
S safeer

Calculate the time taken by the first ball to travel the maximum height:

v=u-gt

v=0 at the maximum height

t=u/g=30/10=3sec

During the three-second of the journey, the ball travels a distance of 

H=ut-0.5gt^2=30\times3-0.5\times10\times9=45m

Let the height of the tower be h. If t is the time taken by the second ball to reach the ground. The total downward journey by the first ball is for (t+4-3) sec, that is t+1 sec

During t+1 sec the first ball travels a distance of 45+h

45+h =0.5 g (t+1)2......(2)

Also 

h= 0.5gt2........(2)   (time taken by second ball)

from 1 and 2

45+h=0.5gt2+gt+0.5g

45+0.5gt2=0.5gt2+10t+5

40=10t

t=4sec

h=0.5 x 10 x 16=80m

The height of tower= 80 m

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