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Asked by Prashant DIGHE | 24 Apr, 2020, 10:00: PM
Let the ball take t seconds to reach ground.

Velocity at (t-1) second = g(t-1)

Distance S travelled in last one second is obtained from the equation of motion ,

S = ut +(1/2)g t2 by substituting t = 1 and intial speed u = g(t-1)

Distance travelled in last second = g(t-1) +(1/2)g = 55

By substituting g = 10 m/s2 , we get from above equation , t = 5 s

Height of tower h = (1/2) g t2 = (1/2)×10×25 = 125 m

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When a thrown object vertically travels equal distance in 5th second and 6th second means , it reaches maximum
height after 5th second and travels downward at 6th second

Hence projection velocity to reach maximum height at t=5 second  , u = g t = 9.8 × 5 = 49 m/s
Answered by Thiyagarajan K | 25 Apr, 2020, 09:06: AM

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