1.

a ball is thrown vertically upward from top of the tower with speed of 100m/s it strike the pond near the base of the tower after 25 sec the height of the tower is ?

Answer»

a ball is thrown vertically upward from TOP of the TOWER with speed of 100m/s it strike the pond near the base of the tower after 25 sec.

initial velocity of ball, u = 100m/s

TIME taken to strike the pond near the base of the tower , t = 25sec.

Let height of tower is H.

when ball is thrown upward, acceleration due to gravity acting on it, due to this after some time ball becomes rest.

let at h height from the top of tower ball becomes rest.

now using formula, v² = u² + 2as

or, 0 = (100)² + 2(-10) × h

or, h = 500m ....(1)

and time taken to reach h = 500m

v = u + at

or, 0 = 100 - 10t => t = 10s

hence, (25 - 10)s = 15s require to fall the ground

now ball is LOCATED at S = (H + 500)m

where velocity of ball , u = 0

and time taken to fall the ground, t = 15s

using formula, s = ut + 1/2 at²

or, -(H + 500) = 0 + 1/2 (-10) × (15)²

or, H + 500 = 5 × 225 = 1125

or, H = 1125 - 500 = 625m

hence, height of tower is 625m


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