Q. A projectile is fired at an angle of $30{^\circ}$ to the horizontal such that the vertical component of its initial velocity is $80\ ms^{- 1}$. Its time of flight is $T$. Its velocity at $t = \frac{T}{4}$ has a magnitude of nearly
A) $200\ ms^{- 1}$
B) $300\ ms^{- 1}$
C) $140\ ms^{- 1}$
D) $100\ ms^{- 1}$
Explanation
(c) $\frac{u_{x}}{u_{y}} = \cot{30{^\circ}} = \sqrt{3}\therefore u_{x} = 80\sqrt{3}\ ms^{- 1}$
$T = \frac{2u_{y}}{g} = \frac{2 \times 80}{10} = 16\ s$
At $t = \frac{T}{4} = 4\ s,v_{x} = 80\sqrt{3}\ ms^{- 1}$
$v_{y} = 80 - 10 \times 4 = 40\ ms^{- 1}$
$\therefore 0\ v = \sqrt{\left( 80\sqrt{3} \right)^{2} + (40)^{2}} = 140\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.