Question

A projectile is fired at 30^o to the horizontal. The vertical component of its velocity is 80 ms^–1. Its time of flight is T. What will be the velocity of the projectile at t = T/2

• A. 80 ms^–1
• B. $80\sqrt{3}$ms^–1
• C. (80/$\sqrt{3}$) ms^–1
• D. 40 ms^–1

Answer 1

Answer: (B)

$80\sqrt{3}$ms^–1

Answer 2

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 $= u\sin\theta = 80$ ⇒ $u = \frac{80}{\sin 30^{o}} = 160m/s$

∴ $u_{horizontal} = u\cos\theta = 160\cos 30^{o} = 80\sqrt{3}m/s.$