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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 |
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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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