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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Physics
Chapter
Motion in Two Dimensions

JEE Advanced 2018 · Paper 2 · Physics · Motion in Two Dimensions

A ball is projected from the ground at an angle of 45 ^ o with the...

A ball is projected from the ground at an angle of 45o{45}^{o} with the horizontal surface. It reaches a maximum height of 120m120m 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 30o{30}^{o} with the horizontal surface. The maximum height it reaches after the bounce (in meters) is _____________.A ball is projected from the ground at an angle of 45 ^ o with the horizontal surface. It reaches a maximum height of

Correct answer

0

Step-by-step solution

A ball is projected from the ground at an angle of 45 ^ o with the horizontal surface. It reaches a maximum height of
H1=u2sin⁡245o2g=120{H}_{1} = \dfrac{{u}^{2}\sin^{2} {45}^{o}}{2g} = 120
⇒u24g=120...(i)⇒ \dfrac{{u}^{2}}{4g} = 120 . . . \left(i\right)
when half of kinetic energy is lost v=u2v = \dfrac{u}{\sqrt{2}}
H2=(u2)2sin⁡230°2g=u216g...(ii){H}_{2} = \dfrac{{\left(\dfrac{u}{\sqrt{2}}\right)}^{2}\sin^{2}30 °}{2g} = \dfrac{{u}^{2}}{16g} . . . \left(ii\right)
From \left(i\right)and (ii)and \left(ii\right),
H2=H14=30m{H}_{2} = \dfrac{{H}_{1}}{4} = 30m
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