A ball is droped from a high rise platform t=0 starting from rest. After 6s another ball is thrown downwards from the same platform with a speed v. The two balls meet at t=18s. What is the value of v?
A
75 m/s
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B
64 m/s
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C
84 m/s
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D
94 m/s
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Solution
The correct option is A 75 m/s
The distance traveled by first ball in t=18s is h1=ut+12gt2=0(18)+0.5(10)(18)2=1620
As the second ball travels same distance h1 in time t=(18−6)=12s, so
1620=vt+0.5gt2 or 1620=12v+0.5(10)(12)2 or 12v=900