Question

A body is projected at an angle of $45^{\circ }$ from a point on the ground at a distance of $30m$ from the foot of a vertical pole of height $20m$. The body just crosses the top of the pole and strikes the ground at a distance $s$ from the foot of the pole on the other side of the pole. Then, $s$

• A. 20 m
• B. 30 m
• C. 50 m
• D. 60 m

Answer 1

Answer: (D)

According to the question, projection of a body shown in the figure below,

Now, from the above figure, let $t$ be the time of crossing of pole,
then, $30=tu\cos 45^{\circ }\dots (i)$
and $20=u\sin 45^{\circ }t-\frac{10}{2}{t}^2\dots (ii)$
Hence, $30\sqrt{2}=ut$ and
$\Rightarrow 20=\frac{u}{\sqrt{2}}t-5t^2$
$\Rightarrow t=\sqrt{2}s$ and $u=30m/s$
Hence, $R=\frac{u^2\sin 2\theta }{g}=\frac{30\times 30\times \sin 90^{\circ }}{10}=90m$
$\therefore$ Distance between pole and the point at which body strike on the ground,
$\Rightarrow s=(90-30)=60m$