Q. A particle is projected with speed $v$ at an angle $\theta\left( 0 < \theta < \frac{\pi}{2} \right)$ above the horizontal from a height $H$ above the ground. If $v =$ speed with which particle hits the ground and $t =$time taken by particle to reach ground, then
A) As $\theta$ increases, $v$ decreases and $t\ $increases
B) As $\theta$ increases, $v$ increases and $t\ $increases
C) As $\theta$ increases, $v$ remains same and $t\ $increases
D) As $\theta$ increases, $v$ remains same and $t\ $decreases
Explanation
(c) From figure,
[image:4_image170.emf]
$H = \left( - v_{0}\sin\theta \right)t + \frac{1}{2}gt^{2}$
$v_{x} = v_{0}\cos\theta$
$v_{y}^{2} = \left( v_{0}\sin\theta \right)^{2} + 2gH$
$v = \sqrt{v_{x}^{2} + v_{y}^{2}}$ at ground
$v = \sqrt{v_{0}^{2} + 2gH}$
It means speed is independent of angle of projection
Also, $\frac{1}{2}gt^{2} = H + t\ v_{0}\sin\theta$
From this, where $\theta$ increases, $t$ increases
Hence, (c) is correct
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.