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A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120m 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° with the horizontal surface. The maximum height it reaches after the bounce, in meters, is____________.

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answer is 30.00.

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Detailed Solution

H1=u2 sin2 452g=120 ⇒u24g=120 ……..(i)When half of kinetic energy is lost means V= u2 H2=u22sin2302g=u216g……..(ii) From (i) & (ii)H2 = 30.00m

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A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120m 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° with the horizontal surface. The maximum height it reaches after the bounce, in meters, is____________.