Physics - Kinematics Question with Solution | TestHub

The ball will be vertical at the highest point so

$t_{up}= \frac{usin\mathrm{\theta }}{g}=2 \Rightarrow usin\mathrm{\theta }=20\mathrm{ }\mathrm{ }\mathrm{ }..\mathrm{ }\dots \dots .(1)$

$\because \mathrm{ta}\mathrm{n}\mathrm{\alpha }=\frac{u\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{\theta }-gt}{u\mathrm{ }\mathrm{c}\mathrm{o}\mathrm{s}\theta }$

$\Rightarrow 1=\frac{20-g\left(1\right)}{u\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{\theta }}\Rightarrow u\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{\theta }=10$… . .(2)

Squaring and adding equation (1) & equation (2), we get

${\left(usin\mathrm{\theta }\right)}^2+{\left(\mathit{\text{u}}\text{cos}\theta \right)}^2={20}^2+{10}^2$

$\Rightarrow {\mathit{u}}^2=500$

$\Rightarrow u=10\sqrt{5} \mathrm{m} {\mathrm{s}}^{-1}$