Question

A projectile is fired at an angle ${30}^{\circ }$ of to the horizontal such that the vertical component of its initial velocity is 80 m/s. Its time of flight is T. Its velocity at t= T/4 has a magnitude of nearly:

• A. 200 m/s
• B. 300 m/s
• C. 150 m/s
• D. 100 m/s

Answer 1

The correct option is C 150 m/s

Vertical component of initial velocity: $u_y=80m/s$

Time of flight is T

$t=\frac{T}{4}=\frac{2u_y\sin \theta }{g}\times \frac{1}{4}$

$=\frac{2\times 80\times \sin 30}{10}\times \frac{1}{4}=2s$

$v_y=u_y-gt$

$=80-10\times 2=80-20=60m/s$

$u\sin {30}^{\circ }=80$

$u=160m/s$

$u_x=u\cos {30}^{\circ }=160\times \frac{\sqrt{3}}{2}=80\sqrt{3}m/s$

At $t=\frac{T}{4},v_x=u_x$ but $v_y$ becomes 60 m/s

$v(t=\frac{T}{4})=\sqrt{v_x^2+v_y^2}$

$=\sqrt{(6400\times 3)+3600}$

$\sqrt{22800}\cong 150m/s$