Q12 of 35 Page 55

A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 s another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v?

Consider Ball 1:

At t = 0, Ball 1 is thrown downward with an initial velocity of 0.


Symbols Used:


u = Initial Velocity; v = Final Velocity; s = Distance Travelled; a = Acceleration


For Ball 1:


u = 0


t = 18 – 0 = 18 s


s = ?


a = g (Acceleration due to gravity)


Now, we know from the equations of motion that:




s = 1587.6 m


Now, for Ball 2:


u = v


t = 18 – 6 = 12 s (Since Ball 2 is dropped at t = 6 s)


s = same as for Ball 1, since both meet at same point at t = 18 s


a = g


Now, using equations of motion:




v = 7.35 ms–1


Hence, the speed of Ball 2 must be 7.35 ms–1 to meet Ball 1 after 18 s.


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