Question

Single Correct Answer Type
If a stone has 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 speed $u$ if the stone is thrown at an angle $\theta$?

• A. $\frac{g}{\cos \theta }\sqrt{\frac{d}{\left(2\right(d\tan \theta -h)}}$
• B. $\frac{d}{\cos \theta }\sqrt{\frac{g}{\left(2\right(d\tan \theta -h)}}$
• C. $\sqrt{\frac{g{d}^2}{h{\cos }^2\theta }}$
• D. $\sqrt{\frac{g{d}^2}{(d-h)}}$

Answer 1

The correct option is B $\frac{d}{\cos \theta }\sqrt{\frac{g}{\left(2\right(d\tan \theta -h)}}$

$u_X=u\cos \theta ,x=d$

$u_Y=u\sin \theta ,y=h$

$d=u\cos \theta .t$

$t=\frac{d}{u\cos \theta }$

$h=u\sin \theta t-\frac{1}{2}g{t}^2$

$=u\sin \theta \frac{t}{u\cos \theta }-\frac{g}{2}\times \frac{d^2}{u^{2}co{s}^2\theta }$

$=d\tan \theta -\frac{g{d}^2}{2u^2{\cos }^2\theta }$

$\frac{g{d}^2}{2u^{2}co{s}^2\theta }=d\tan \theta -h$

$u=\frac{d}{\cos \theta }\sqrt{\frac{g}{2(d\tan \theta -h)}}$