Math, asked by ItzSavageGirlIsha, 3 days ago

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?

(a) T1+T2/2
(b) T2-T1/2
(c) √T1T2
(d) T1+T2

Solve this Question !!


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Answers

Answered by fakekuldeep16
1

i hope it's help you thank you

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Answered by s8a1583aritra1756
2

≡hey Isha di....

≡ hru

≡mine all good:)

≡ here is your answer....

Correct option is C)

Given :-

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

To Find :-

In what time the ball shall reach the ground if it is allowed to fall freely under gravity from top of the tower?

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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