Question

A body is projected at an angle of $45^{\circ}$ from a point on the ground at a distance of $30 \,m$ from the foot of a vertical pole of height $20\, m$. 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=t \,u \cos\, 45^{\circ}\,\,\,\,\,\dots(i)$
and $ 20=u \sin 45^{\circ} t-\frac{10}{2} t^{2}\,\,\,\,\,\dots(ii)$
Hence, $30 \sqrt{2}=u t$ and
$\Rightarrow \, 20=\frac{u}{\sqrt{2}} t-5 t^{2}$
$\Rightarrow \, t=\sqrt{2} s$ and $u=30\, m / s$
Hence, $R=\frac{u^{2} \sin 2 \theta}{g}=\frac{30 \times 30 \times \sin 90^{\circ}}{10}=90\, m$
$\therefore $ Distance between pole and the point at which body strike on the ground,
$\Rightarrow \, s=(90-30)=60\, m$