From top of a tower 100m height,a ball is dropped and same time another ball is projected upward from ground with velocity 25m/s.Find when and where two balls meet.
Asked by sagarikadevnath devnath | 1st Oct, 2013, 06:58: PM
Expert Answer:
If the two balls meet at a distance x from the bottom of the tower.
(100 - x) = 1/2gt2
.(1)
and h = ut + 1/2at2
so, x = 25t (1/2)gt2
.(2)
From eqn (1) & (2),we get,
100 = 25t
t = 4sec
From eqn (2)
x = 25 *4 (1/2)(9.8)(4)2
x = 21.6m
Therefore, the height at which the two balls will meet = 21.6m and time, t=4sec.
If the two balls meet at a distance x from the bottom of the tower.
(100 - x) = 1/2gt2 .(1)
and h = ut + 1/2at2
so, x = 25t (1/2)gt2 .(2)
From eqn (1) & (2),we get,
100 = 25t
t = 4sec
From eqn (2)
x = 25 *4 (1/2)(9.8)(4)2
x = 21.6m
Therefore, the height at which the two balls will meet = 21.6m and time, t=4sec.
Answered by Faiza Lambe | 1st Oct, 2013, 10:03: PM
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