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Q.

If a stone is to hit 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 launched at an angle $θ$ ?

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a

$\frac{g}{\cos θ} \sqrt{\frac{d}{2 (dtan θ - h)}}$

b

$\frac{d}{\cos θ} \sqrt{\frac{g}{2 (dtan θ - h)}}$

c

$\sqrt{\frac{{gd}^{2}}{{hcos}^{2} θ}}$

d

$\sqrt{\frac{{gd}^{2}}{(d - h)}}$

answer is B.

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Detailed Solution

$\begin{array}{l} h = (usin θ) t - \frac{1}{2} {gt}^{2} \\ d = (ucos θ) t \end{array}$
or $t = \frac{d}{ucos θ}$
$∴ h = usin θ \cdot \frac{d}{ucos θ} - \frac{1}{2} g \cdot \frac{d^{2}}{u^{2} \cos^{2} θ}$
$∴ u = \frac{d}{\cos θ} \sqrt{\frac{g}{2 (dtan θ - h)}}$ .

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