Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics2-D & 3DEasy2 minPYQ_2020
PhysicsEasynumerical

A ball is projected from the ground at an angle of45owith the horizontal surface. It reaches a maximum height of120 mand 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 of30owith the horizontal surface. The maximum height it reaches after the bounce (in meters) is _____________.

Answer:
30.00
Solution:



H1=u2sin2 45 o 2g=120

     u24g=120 .....(i)

when half of kinetic energy is lostv=u2

H2=u22sin2 30 o 2g=u216g......(ii)

From (i) and (ii)

H2=H14=30 m on 30.00

Subject:PhysicsTopic:KinematicsSubtopic:2-D & 3D
2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...