A body is projected at an angle of $45^{\circ}$ from a point on the ground at a distance of $30 \mathrm{~m}$ from the foot of a vertical pole of height $20 \mathrm{~m}$. 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…

Question

A body is projected at an angle of $45^{\circ}$ from a point on the ground at a distance of $30 \mathrm{~m}$ from the foot of a vertical pole of height $20 \mathrm{~m}$. 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, $s$

Aicharya · Accepted Answer

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, $30=t u \cos 45^{\circ}$ ...(i) and $20=u \sin 45^{\circ} t-\frac{10}{2} t^2$ ...(ii) Hence, $30 \sqrt{2}=u t$ and $\begin{array}{ll} \Rightarrow & 20=\frac{u}{\sqrt{2}} t-5 t^2 \\ \Rightarrow \quad & t=\sqrt{2} \mathrm{~s} \text { and } u=30 \mathrm{~m} / \mathrm{s} \\ \text {Hence, } R= & \frac{u^2 \sin 2 \theta}{g}=\frac{30 \times 30 \times \sin 90^{\circ}}{10}=90 \mathrm{~m} \end{array}$ $\therefore$ Distance between pole and the point at which body strike on the ground, $\Rightarrow \quad s=(90-30)=60 \mathrm{~m}$ So, the correct option is (d).

A body is projected at an angle of \(45^{\circ}\) from a point on the ground at a distance of \(30…

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