From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in________s.
Text Solution
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(3) Let height of tower be h and speed of projection in first two cases be u.
For case-I : 2 nd equation s = ut + 
at 2 
....(i)
For case-II: h = u(l.5) + 
g(l.5)
2
...(ii)
Multiplying equation (ii) by 4 we get
4h = 6u + 4.5 g .... (iii)
equation (i) + equation (iii) we get 5h = 22.5g
h = 4.5g... (iv)
For case-Ill :
h=0 + 
gt 2 ...(v)
Using equation (4) & equation (5)
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