Physics - Kinematics Question with Solution | TestHub

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 $$\begin{array}{rl}& =\mathrm{OA}\tan \theta -\frac{1}{2}\frac{\mathrm{g}(\mathrm{OA}{)}^2}{{\mathrm{u}}^2{\cos }^2\theta }\\ & =4-\frac{1}{2}\times \frac{10\times 16}{10\times 10\times \frac{1}{\sqrt{n}}\times \frac{1}{\sqrt{0}}}=2.4\mathrm{m}\end{array}$$