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Question

A ball thrown vertically upwards at a certain speed from the top of a tower of height 50 m m 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 s
Taking 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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