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Q.

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?(take g = 10 ms-2)

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a

74 ms-1

b

55 ms-1

c

40 ms-1

d

60 ms-1

answer is A.

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Detailed Solution

For first ball, u = 0∴s1 = 12gt12  =12×g×182For second ball, initial velocity = v∴s2 = vt2+12gt22t2 = 18-6 = 12 s⇒ s2 = s2 = v×12+12g(12)2   Here, s1 = s212g(18)2 = 12v+12g(12)2    ⇒ v = 74ms-1
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