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Question

Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t=0 s. Ball B is thrown vertically down with an initial velocity u at t=2 s. After a certain time, both balls meet 100 m above the ground. Find the value of u in ms1 [ use g=10 m/s2 ]

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Solution


Let us assume that they meet at t=t0

A:80=12gt20 ....(i)

B:80=u(t02)+12g(t02)2 ....(ii)

From (i), t0=4

80=2u+5(2)2

u=30 m/s

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