Question

A ball is thrown upward with certain velocity. Another
ball is released from a height of 60 m. They meet at
a height of 15m. Then what is the initial velocity of
the 1st ball (g = 10 ms)
(a)20m/sec
(b)30m/sec
(c)40m/sec
(d)60m/sec​

Answer 1

Answer:

(a) 20m/s

Explanation:

Distance travelled by 2nd ball=60-15=45m

s=ut+1/2at^2

45=1/2*10*t^2

45=5t^2

t=3second

Therefore,they collide after 3sec

s=ut +1/2at^2

15=3u-5*9

15=3u-45

Therefore, u=20m/s

Answer 2

Answer:

Answer is a

20 m/ sec

Hope it will help yu