A body thrown vertically up with a velocity ‘U’. It reaches the maximum height ‘H’ in ‘t’ seconds. Then correct statement among the following

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At a height $\frac{H}{2}$ from ground its velocity is $\frac{' U'}{2}$

At a time 2t its velocity is ‘-U’

At a time 2t its velocity is ‘-6U’

At time t its velocity is ‘U’

answer is C.

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Detailed Solution

a) At half of maximum height from $V^{2} - U^{2} = 2 a s, h = \frac{U^{2}}{2 g}$
$\begin{array}{l} \Rightarrow V^{2} - U^{2} = 2 g (\frac{h}{2}) \Rightarrow V^{2} - U^{2} = - g (\frac{U^{2}}{2 g}) \\ \Rightarrow V^{2} - U^{2} = \frac{- U^{2}}{2} \Rightarrow V^{2} = \frac{- U^{2}}{2} + U^{2} \Rightarrow V = \frac{U}{\sqrt{2}} \end{array}$
b) At time ‘t’ the body is in maximum height its velocity is zero. ‘t’ is time of ascent $t = \frac{U}{g}$
c) At time ‘2t’ from $V = U + a t t = \frac{U}{g}, a = - g$
$V = U + 2 a t \Rightarrow V = U + 2 (- g) \frac{U}{g}$
$V = U - 2 U \Rightarrow V = - U$