A stone is released from rest from the top of a tower. Exactly at the same time another stone is thrown straight vertically up with a speed of 25 m/s from the bottom of the tower. The two stones just cross each other after 2 seconds. (a) What is the height of the tower (b) What is the distance separating the two stones at the time the second stone reaches the maximum height ? Take g = 10 m/s2. [ Ans. (a) 50 m (b) 12.5 m ]
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Explanation:
Let h be the height if tower.
s₁ = dist covered by 1st stone in time 2 sec
= ut + 1/2 gt²
= 1/2 x 10 x 4
= 20m
s₂ = 25x2 -1/2x10x4
= 50 - 20
= 30 m
Height of tower h = s₁ + s₂
h = 20 + 30
= 50 m
Time taken by 2nd stone to reach max ht is
t =u/g = 25/10 = 2.5 s
distance travelled by 1st = 1/2 x 10x 6.25
= 31.25 m
= 18.75 m from bottom
dist travelled by 2nd = 25x2.5 -1/2 x10 x6.25
=62.5 -31.25
= 31.25 m from bottom
dist of separation = 31.25 - 18.75 = 12.5 m
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