Question

A ball is thrown upward with a velocity of 20 m/s from top of a tower having a height of 50 m. What time elap
when it again passes the edge of tower ? (g = 10 ms?)
(a) 2
(b) 4 s
(c) 8 s
(d) 5 s​

Answer 1

Answer:

4 sec

Explanation:

Given ,

initial velocity = 20 m/s

a = g = 10 m/s²

Maximum height is given by ,

H = u² / 2g

   = 400 / 20

   = 20 m

time is given as ,

t = v-u / a

 = u/g

 = 20 / 10

 = 2 sec

the ball takes 2 seconds to reach maximum height .

While falling freely :-

The height will be same ie 20 m

H = 1/2gt²

 20  = 1/2 × 10 × t²

20 = 5t²

t² = 20 / 5

   = 4

t = √4

 = 2 sec

The ball takes two seconds to fall freely.

Therefore , the total time taken for the ball to reach the edge of the tower will be 4 sec  [ as time taken to reach maximum height is 2 sec and to fall freely till the edge of the tower is 2 sec ]

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