Physics - Kinematics Question with Solution | TestHub

Given,
Initial velocity of projectile, $u=40\mathrm{m}/\mathrm{s}$
Angle, $\theta =30^{\circ }$
Time of flight
$\mathrm{T}=\frac{2\mathrm{u}\sin \theta }{\mathrm{g}}=\frac{2\times 40\times 1}{10\times 2}=45\left(\because \mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2\right)$
It means projectile is at maximum height at $t=$ $2\mathrm{s}$. At maximum height vertical component of velocity is zero.
$\begin{array}{rl}& \text{ Velocity at }\mathrm{t}=2\mathrm{s}={\mathrm{V}}_{\mathrm{x}}=\mathrm{u}\cos \theta =40\cos 30^{\circ }\\ & =20\sqrt{3}{\mathrm{ms}}^{-1}\text{. }\\ & \end{array}$