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A man is standing on top of a building 100 m high. He throws two ball vertically, one at `t=0` and after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. At `t=2`, both the balls reach to their and second ball is `+15m`.
Q. The speed of first ball is

Text Solution

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Let the first ball be thrown with vleocity (u) and second ball with a time interv al (x) Taking vertical motion of ball first for time (t = x), we have
S_1 =u (t+ x) - 1/2 g (t + x)^2` …(i) `
Taking upaward motion of second ball for time (t), we have
` S_2 = u/2 t 1/2 gt^2` ....(ii)
:. S_1 - s_2 = u (tt+ x - t/2 g (t^2 + 2 xt = x^2 - t^2)`
`= u (t/2 + x) - 1/2 g ( 2 xt + x^2)`
As per question, `S_1 -S_2 = 15 m , t=2 s and x = 1 s [ as x lt 2 s]`
Then, ` 15 =u (2/2 + 1) - 1/2 xx 10 (2xx 1 xx 2 1^2)`
or ` 15 =u xx 2 - 25`
or ` u = 40 //2 = 20 m//s`
The velocity other ball ` = u/ 2 = (20)/2 = 10 m//s`. Time interval ` x = 1 s`
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