1.

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

Answer»

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



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