Physics - Kinematics Question with Solution | TestHub

After $2 \mathrm{s}$ the velocity is $20 \mathrm{m} {\mathrm{s}}^{-1}$

$\Rightarrow v_x^2+v_y^2=v^2=400$

$\Rightarrow u^2{\cos }^{2}45°+{\left(u\sin 45°-gt\right)}^2=\frac{u^2}{2}+{\left(\frac{u}{\sqrt{2}}-20\right)}^2=400$

$\Rightarrow \frac{u^2}{2}+\frac{u^2}{2}+400-\frac{40}{\sqrt{2}}u=400$

 $\Rightarrow u^2=\frac{40}{\sqrt{2}}u\Rightarrow u=\frac{40}{\sqrt{2}} \mathrm{m} {\mathrm{s}}^{-1}$

Now the maximum height of the projectile is given by,

$$\begin{gathered} H=\frac{u^2{\sin }^2\theta }{2g} \\ \Rightarrow H=\frac{1600}{20\times 2\times 2}=20 \mathrm{m} \end{gathered}$$