Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics2-D & 3DMedium2 minQB

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The velocity at the maximum height of a projectile is half of its initial velocity $u$ . Its range on the horizontal plane is:

Answer:

Solution:

At the highest point only the horizontal velocity remains.

$u cos θ = \frac{u}{2} \Rightarrow θ = 6 0^{\circ}$

$R = \frac{u ^{2} sin ( 2 θ )}{g}$

$R = \frac{u ^{2} sin 12 0 ^{\circ}}{g} = \frac{3 u ^{2}}{2 g}$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:2-D & 3D

⏱ 2mℹ️ Source: QB

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