A ball of mass 0.5 kg , initially at rest on the ground, is hit vertically upwards by a bat. The upward force exerted by the bat on the ball increases linearly from zero to a maximum value of 100 N in 0.05 s , and then decreases linearly to zero in the next 0.05 s . Assuming the effect of gravity is negligible during the impact, the maximum height reached by the ball is (Take g = 10 m s ⁻² ):

Step-by-step solution

The impulse imparted to the ball is equal to the area under the Force-Time graph. Since the force increases and then decreases linearly, the graph forms a triangle with base t = 0.05 + 0.05 = 0.1 s and height F_ max = 100 N . Impulse, J = 1 2 base height = 1 2 0.1 100 = 5 N s . According to the impulse-momentum theorem, J = p = m v - 0 . 5 = 0.5 v v = 10 m s ⁻¹ . The maximum height reached by the ball is given by kinematics: h = v^2 2g = (10)^2 2 10 = 100 20 = 5 m . Answer: 5 m