Question 222 of 308

A projectile fired at $30^{\circ}$ to the ground is observed to be at same height at time $3s$ and $5s$ after projection, during its flight. The speed of projection of the projectile is $m{s}^{-1}$.
(Given $g=10m{s}^{-2}$)

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Answer: 80

The time taken to reach maximum height is
$T'=\dfrac{u\sin \theta }{g}$.
The total time of flight is $T=(3+5)s=8s$.
Hence,
$T'=\left(\dfrac{T}{2}\right)=\left(\dfrac{3+5}{2}\right)=\left(\dfrac{u\sin \theta }{g}\right)$
$\Rightarrow 4=\dfrac{(u)\times \dfrac{1}{2}}{10}$
$\Rightarrow u=80m{s}^{-1}$