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# Solve this

Asked by venkat 25th December 2017, 8:56 PM
Since Q is projected after 1 second, P will reach the maximum height and meet Q when it is returning back.

Let maximum height reached by P is H.

H = (u×u)  /(2×g), where u is initial projection velocity 30 m/s

H = (30×30) / (2*10) = 45 m.

after reaching the maximumheight let P travels back for t seconds and meet Q.

Let the distance travelled by P in that t seconds is h and it is given by

h = (1/2)g×t2 = (1/2)×10×t2 = 5×t2 .........................(1)

Then distance travelled by Q when it meets P is 45-h and the time taken is (2+t)s. This distance is given by

45-h = 30×(2+t) - (1/2)×10×(2+t)2

after substituting for h from eqn.(1), we will get

45 - 5×t2 = 30×(2+t)-5×(2+t)2

solving the above eqn for t, we get t = 0.5 s.

Hence P travelled 3.5 s before meetin Q
Answered by Expert 27th December 2017, 3:13 PM
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