First slide

Projectile motion

Question

A projectile is thrown in the upward direction making an angle of 60^o with the horizontal direction with a velocity of 150 ms^-1. Then the time after which its inclination with
the horizontal is 45^o is

Moderate

A

$15 (\sqrt{3} - 1) s$
B

$15 (\sqrt{3} + 1) s$
C

$7.5 (\sqrt{3} - 1) s$
D

$7.5 (\sqrt{3} + 1) s$

Solution

At the two points of the trajectory during projectile motion, the horizontal component of the velocity is same. Then
$150 \times \frac{1}{2} = v \times \frac{1}{\sqrt{2}} or v = \frac{150}{\sqrt{2}} {ms}^{- 1}$
$Initially: u_{y} = u \sin 60^{\circ} = \frac{150 \sqrt{3}}{2} {ms}^{- 1}$
$Finally: v_{y} = v \sin 45^{\circ} = \frac{150}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{150}{2} {ms}^{- 1}$
$But v_{y} = u_{y} + a_{y} t or \frac{150}{2} = \frac{150 \sqrt{3}}{2} - 10 t$
$10 t = \frac{150}{2} (\sqrt{3} - 1) or t = 7.5 (\sqrt{3} - 1)$

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