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Physics - Kinematics Question with Solution | TestHub

PhysicsKinematicsRelativeMedium2 minQB

PhysicsMediumsingle choice

The horizontal range of a projectile is $R$ and the maximum height attained is $H$ . A wind gives the projectile a constant horizontal acceleration $g /4$ in the direction of motion. Under the same conditions of projection, the horizontal range will now be:

Options:

Answer:

Solution:

Extra distance = $\frac{1}{2} \cdot \frac{g}{4} \cdot T^{2} = \frac{u ^{2} sin ^{2} θ}{2 g} = H$ . Hence $R^{'} = R + H$ . Option B.

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:Relative

⏱ 2mℹ️ Source: QB

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