1. Tardigrade
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  4. 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

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Q. A projectile is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal direction with a velocity of $150\, ms ^{-1}$. Then the time after which its inclination with the horizontal is $45^{\circ}$ is

Motion in a Plane

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)$