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
- Au^2 + v^2 2g
- Bu^2 + v^2 4g
- Cu^2 - v^2 4g
- 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