Question

The figure shows two positions A and B at the same height h above the ground. If the maximum height of the projectile is H, then determine the time t elapsed between the positions A and B in terms of H.

• A. $\sqrt{\frac{4}{g}(H-h)}$
• B. $\sqrt{\frac{2}{g}(H-h)}$
• C. $\sqrt{\frac{8}{g}(H-h)}$
• D. $\sqrt{\frac{6}{g}(H-h)}$

Answer 1

The correct option is C

$\sqrt{\frac{8}{g}(H-h)}$

Let T be the time of flight.
We know that, $T=\frac{2usin\theta }{g}\Rightarrow T^2=\frac{4u^{2}si{n}^2\theta }{g^2}=\frac{8}{g}\left(\frac{u^{2}si{n}^2\theta }{2g}\right)=\frac{8H}{g}$
In similar way, we can write $t^2=\frac{8(H-h)}{g}$
$\therefore t=\sqrt{\frac{8}{g}(H-h)}$