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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 by it is $H$ . A strong wind now begins to blow in the direction of motion of the projectile, giving it a constant horizontal acceleration $g /2$ . Under the same conditions of projection, the horizontal range of the projectile will now be:

Options:

Answer:

Solution:

The time of flight is unchanged. Extra distance = $\frac{1}{2} \cdot \frac{g}{2} \cdot T^{2} = \frac{g}{4} \cdot \frac{4 u ^{2} sin ^{2} θ}{g ^{2}} = \frac{u ^{2} sin ^{2} θ}{g} = 2 H$ . Hence $R^{'} = R + 2 H$ . Option D.

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:Relative

⏱ 2mℹ️ Source: QB

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