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Question

# A ball thrown vertically upwards at a certain speed from the top of a tower, reaches the ground after 9 seconds. Another ball thrown vertically downwards with same speed from the same tower reaches the ground in 4 seconds. How much time will the ball take if it is just dropped from the tower?

A
6 s
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B
7 s
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C
5 s
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D
8 s
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Solution

## The correct option is A 6 sTaking upward direction as (+ve) and downward direction as (-ve) y-axis. For the ball in the first case, s=−h,a=−g,t=t1,u=u0 Applying the equation of motion for first case, we get, −h=u0t1+12(−g)t21 Multiplying this whole equation by t2, we get, −ht2=u0t1t2+12(−g)t21t2 ...(i) For the ball in the second case, s=−h,a=−g,t=t2,u=−u0 . Simillarly for second case, −h=(−u0)t2+12(−g)t22 Multiplying this whole equation by t1, we get, −ht1=(−u0)t2t1+12(−g)t22t1 ....(ii) When the ball is just dropped, −h=12(−g)t2 ⇒h=12(g)t2 ....(iii) Adding the equations (i) and (ii), we get, −h(t1+t2)=−12gt1t2(t1+t2) ⇒h(t1+t2)=12gt1t2(t1+t2) ⇒h=12gt1t2 ...(iv) Comparing equations (iii) and (iv), we get, 12gt2=12gt1t2 t=√t1t2 t=√9×4=√36=6 s

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