First slide

Projection Under uniform Acceleration

Question

A projectile is thrown in the upward direction making an angle of 60° with the horizontal direction with a velocity of 150 ms^-1. Then the time after which its inclination with the horizontal is 45° 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
$150 \times \frac{1}{2} = v \times \frac{1}{\sqrt{2}} or v = \frac{150}{\sqrt{2}} {ms}^{- 1} Initially: u_{y} = usin 60 ° = \frac{150 \sqrt{3}}{2} {ms}^{- 1} Finally: v_{y} = vsin 45 ° = \frac{150}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{150}{2} {ms}^{- 1} But vy = {uy}_{y} + ayt or \frac{150}{2} = \frac{150 \sqrt{3}}{2} - 10 t ∣ 10 t = \frac{150}{2} (\sqrt{3} - 1) or t = 7.5 (\sqrt{3} - 1)$

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