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JEE MainPhysicsMotion in One Dimension

A ball is projected vertically upwards with an initial speed u from the ground. It experiences a constant air resistance throughout its motion and returns to the ground with a speed v . If g is the acceleration due to gravity, the maximum height reached by the ball is :

Options

  1. Au^2 + v^2 2g
  2. Bu^2 + v^2 4g
  3. Cu^2 - v^2 4g
  4. Du^2 - v^2 2g

Correct answer

B. u^2 + v^2 4g

Step-by-step solution

Let h be the maximum height reached and a_r be the constant deceleration due to air resistance. During the upward journey, the net acceleration is g + a_r (downwards). Using the kinematic equation v^2 = u^2 + 2as for the ascent (final speed is zero): 0 = u^2 - 2(g + a_r)h 2(g + a_r)h = u^2 ... (i) During the downward journey, the net acceleration is g - a_r (downwards). Using the same kinematic equation for the descent (initial speed is zero): v^2 = 0 + 2(g - a_r)h 2(g - a_r)h = v^2 ... (ii) Adding equations (i) an

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