Question

Q. A ball thrown upward from the top of tower with speed V reaches the ground in T1 second. If this ball is thrown downward from the top of the same tower with speed V it reaches the ground in T2 second. In what time the ball shall reach the ground if it is allowed to fall freely under gravity from top of the tower?
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Answer 1

Answer:

Solution :-

As it is given that the speed is V. So, By using second equation of kinematics

s = ut + 1/2 at²

But here,

a = g

u = V

s = S

S = Vt₂ + 1/2 × g × (T2)²

Since, the ball is thrown upward. So, time taken when they will come at same point = T1 - T2

Time taken by T1 for reaching top

T1/2

Time taken by T2 for reaching top

T2/2

Time taken to come at top including both upward and downward\

T1/2 - T2/2

T1 - T2/2

As when it come at top. Then, its velocity will become 0

S = Vt₂ + 1/2 × g × (T2)²

0 = V - g × (T1/2- T2/2)

0 = V - g × (T1 - T2/2)

0 + V = g(T1 - T2/2)

V = g(T1 - T2/2)

Now

S = T2g(T1 - T2/2) + 1/2 × g × T2²

S = T2g(T1 - T2/2) + g/2 × T2²

Taking g and T2 as common

S = g × T1 × T2/2 (1)

S = g × T²/2

S = 1/2 × g × T²

S = 1/2 gT² (2)

On comparing both

g × T1 × T2/2 = 1/2 × gT²

g × T²/2 = 1/2 × gT²

T²/2 = 1/2 × T²

T² = T²

T = √(T1T2)

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Answer 2

Answer:

h=−vt1+21gt12,h=vt2+21gt22

Now, t1h+t2h=21g(t1+t2) or h=21gt1t2

Also, h=21gt2∴t2=t1t2or t=t1t2.