JEE Advanced 2018 · Paper 2 · Physics Q8

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PhysicsNumerical+3Question 8 of 54

A ball is projected from the ground at an angle of $45^\circ$ 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 $30^\circ$ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is ___________.

Your answer

Answer and worked solution

Official answer: 29.6

Before the bounce. $H=\dfrac{u^2\sin^245^\circ}{2g}=\dfrac{u^2}{4g}=120$ , so $u^2=480g$ .

After the bounce. Half the KE is lost, so $v^2=\dfrac{u^2}2=240g$ . The ball leaves at $30^\circ$ :
$H'=\frac{v^2\sin^230^\circ}{2g}=\frac{240g\times\frac14}{2g}=30\text{ m}.$

Answer: 30

Topics:Projectile motion Work-energy theorem

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JEE Advanced 2018 · Paper 2, Physics question 8