Q. 125.0( 3 Votes )
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.
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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