Question

From the top of tower, a stone is thrown up. It reaches the ground in t1 s. A second stone thrown with the same speed reaches the ground in t2 s. A third stone released from rest reaches the ground in t3 s. Then

Answer 1

Explanation:

let height of tower is H, acceleration due to gravity is a, 

case 1) initial velocity is −u (assuming downward direction as positive.

then H=−ut+21at2=>H=−5u+225a

for case 2) initial velocity is u

then  H=ut+21at2=>H=u+21a

multiplying 2nd with 5 and adding both equation we get 

6H=15a=>H=2.5a

now for 3rd case u=0, we get 2.5a=21at2=>t=5sec.

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