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Question

A ball is projected from the ground at an angled of 45o 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 30o with the horizontal surface. The maximum height it reaches after the bounce, (in metres), is

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Solution

Given,
Initial maximum hight obtained by the ball= 120 m.
By using projectile motion equation (maximum hight)
H1=u2sin2452g=120
u24g=120 …. (i)
When half of kinetic energy is lost v = u2

so again , H2=(u2)2sin2302g=u216g...… (ii)

From (i) and (ii)
H2=14[u24g], (where, u24g=120 from eq. (1))
so, 14×(120)=30 m

Hence, the maximum hight after the collision will be 30 m.

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