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 - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

At max height, the final velocity will be 0.

v= 0

v = u + at

$∴ 0 = u + at$

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

$∴ u = gt$ ..eq (1)

$s = ut + \frac{1}{2} {at}^{2}$

Substituting equation 1 in the above equation,

$H = {gt}^{2} - \frac{1}{2} {gt}^{2}$

Where H is the maximum height.

$g= \frac{2 H}{t^{2}}$

Let the velocity at height h be V_h,

$V^{2} = u^{2} + 2 as$

$∴ {V_{h}}^{2} = {(gt)}^{2} - 2 (\frac{2 H}{t^{2}}) h$

$∴ {V_{h}}^{2} = {(\frac{2 H}{t^{2}} t)}^{2} - 2 (\frac{2 H}{t^{2}}) h$

$∴ {V_{h}}^{2} = \frac{4}{t^{2}} (H^{2} - Hh)$

V_h= $\frac{4}{T} \sqrt{(H^{2}} - H h)$

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