JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Kinematics

A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of

120 m

and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of

3 0^{\circ}

with the horizontal surface. The maximum height it reaches after the bounce, in metres, is ____.

Answer: 30

Solution

The initial maximum height is

H_{0} = \frac{( v _{0} s i n 4 5 ^{\circ} ) ^{2}}{2 g} = \frac{v _{0}^{2}}{4 g} = 120

m. After the bounce, the kinetic energy is halved, so

\frac{1}{2} m v^{2} = \frac{1}{2} \cdot \frac{1}{2} m v_{0}^{2}

, giving

v = \frac{v _{0}}{2}

. The new maximum height is

H = \frac{( v s i n 3 0 ^{\circ} ) ^{2}}{2 g} = \frac{v ^{2}}{8 g} = \frac{v _{0}^{2}}{16 g} = \frac{H _{0}}{4} = \frac{120}{4} = 30

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Numerical answer

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JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Kinematics

Solution

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