First slide

Projectile motion

Question

A particle is projected vertically upwards from O with velocity v and a second particle is projected at the same instant from P (at a height h above O) with velocity v at an angle of projection $θ$ . The time when the distance between them is minimum is

Difficult

A

$\frac{h}{2 v \sin θ}$
B

$\frac{h}{2 v \cos θ}$
C

$h / v$
D

$h / 2 v$

Solution

Relative acceleration between the two particles is zero. The distance between then at time t is
$s = \sqrt{\{h - (v - v \sin θ t}^{2} + (v \cos θ t)^{2}}$
$or s^{2} = {h - (v - v \sin θ) t}^{2} + (v \cos θ t)^{2}$
$s is minimum when \frac{d}{d t} (s^{2}) = 0$
$or 2 {h - (v - v \sin θ) t} {v \sin θ - v} + 2 v^{2} \cos^{2} θ t = 0$
$or t = \frac{h}{2 v}$

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