Q.
If a stone $s$ to hit at a point which is at a distance $d$ away and at a height $h$ above the point from where the stone starts, then what is the value of initial sped $u$, if the stone is launched at an angle $Q$?
A) $\frac{g}{\cos\theta}\sqrt{\frac{d}{2(d\tan\theta - h)}}$
B) $\frac{d}{\cos\theta}\sqrt{\frac{g}{2(d\tan\theta - h)}}$
C) $\sqrt{\frac{gd^{2}}{h\cos^{2}\theta}}$
D) $\sqrt{\frac{gd^{2}}{(d - h)}}$
Explanation
(b) $h = \left( u\sin\theta \right)t - \frac{1}{2}gt^{2}$
$d = \left( u\cos\theta \right)t\ \ or\ \ \ t = \frac{d}{u\cos\theta}$
$h = u\sin\theta.\frac{d}{u\cos\theta} - \frac{1}{2}g.\frac{d^{2}}{u^{2}\cos^{2}\theta}$
$u = \frac{d}{\cos\theta}\sqrt{\frac{g}{2(d\tan{\theta - h})}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.