A ball projected from ground at an angle of $45^{\circ}$ just clears a wall in front. If point of projection is $4 \mathrm{~m}$ from the foot of wall and ball strikes the ground at a distance of $6 \mathrm{~m}$ on the other side of the wall, the height of the wall is :

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Aicharya · Accepted Answer

As ball is projected at an angle $45^{\circ}$ to the horizontal therefore Range $=4 \mathrm{H}$ or $10=4 \mathrm{H} \Rightarrow \mathrm{H}=\frac{10}{4}=2.5 \mathrm{~m}$ $(\because$ Range $=4 \mathrm{~m}+6 \mathrm{~m}=10 \mathrm{~m})$ Maximum height, $\mathrm{H}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}$ $\therefore \mathrm{u}^2=\frac{\mathrm{H} \times 2 \mathrm{~g}}{\sin ^2 \theta}=\frac{2.5 \times 2 \times 10}{\left(\frac{1}{\sqrt{2}}\right)^2}=100$ or, $\mathrm{u}=\sqrt{100}=10 \mathrm{~ms}^{-1}$ Height of wall PA $ \begin{aligned} & =\mathrm{OA} \tan \theta-\frac{1}{2} \frac{\mathrm{g}(\mathrm{OA})^2}{\mathrm{u}^2 \cos ^2 \theta} \\ & =4-\frac{1}{2} \times \frac{10 \times 16}{10 \times 10 \times \frac{1}{\sqrt{n}} \times \frac{1}{\sqrt{0}}}=2.4 \mathrm{~m} \end{aligned} $

A ball projected from ground at an angle of $45^{\circ}$ just clears a wall in front. If point of projection…

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