Question

A projectile is fired at ${30}^{\circ }$ to the horizontal. The vertical component of its velocity is $80m{s}^{-1}$. Its time of flight is T. What will be the velocity of the projectile at $t=\frac{T}{2}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

At half of the time of flight, the position of the projectile will be at the highest point of the parabola and at that position particle possess horizontal component of velocity only.

$Given{U}_{vertical}=usin\theta =80\Rightarrow u=\frac{80}{sin{30}^{\circ }}=160m/s\therefore u_{horizontal}=ucos\theta =160cos{30}^{\circ }=80\sqrt{3}m/s$