A small body is thrown vertically up which reaches a maximum height of H and whose total time of flight is T. Its velocity at a height h will be

# A small body is thrown vertically up which reaches a maximum height of H and whose total time of flight is T. Its velocity at a height h will be

1. A

$\frac{4}{T}\sqrt{\left({H}^{2}}-Hh\right)$

2. B

3. C

$\frac{8}{\mathrm{T}}\left(\mathrm{H}-\mathrm{h}\right)$

4. D

$\frac{4}{\mathrm{T}}\left(\mathrm{H}+\mathrm{h}\right)$

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### Solution:

At max height, the final velocity will be 0.

v= 0

v = u + at

Since the acceleration due to gravity is in the opposite direction,
a = -g

..eq (1)

Substituting equation 1 in the above equation,

Where H is the maximum height.

Let the velocity at height h be Vh,

Vh$\frac{4}{T}\sqrt{\left({H}^{2}}-Hh\right)$

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