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.
• B.
• C.
• D.

Answer 1

The correct option is C

Time of flight for this part of motion (between points A and B, both at height h) assuming speed of projection to be $v$ and angle of projection to be $\alpha$

$T=\frac{2vsin\alpha }{g}$ ---------(I)

Now we have to eliminate $v$ & $\alpha$ as these variables we assumed.

Hmmmmm..........How do we do that, okay what else is given?

Yes, the maximum height i.e. $(H-h)$.

$\Rightarrow$ $(H-h)$ = $\frac{v^{2}si{n}^2\alpha }{2g}$ ----------(II)

Thank goodness!

Now from equation (I) $\Rightarrow$ $vsin\alpha$ = $\frac{Tg}{2}$

Equation (II) $\Rightarrow$ $(H-h)$ = $\frac{T^2{g}^2}{4\left(2g\right)}$

$\Rightarrow$$(H-h)$ = $\frac{g{T}^2}{8}$

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