ExamBro
ExamBro
JEE Advanced · Physics · 4. Motion in 2D

A ball is projected from the ground at an angle 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 meters) is _____________.

  1. A 32
  2. B 30
  3. C 34
  4. D 36
Verified Solution

Answer & Solution

Correct Answer

(B) 30

Step-by-step Solution

Detailed explanation


H1=u2sin2 45 o 2g=120
u24g=120 ...i
when half of kinetic energy is lost v=u2
H2=u22sin2 30°2g=u216g ...ii
From i and ii,
H2=H14=30 m
Same subject
Explore more questions on app
From JEE Advanced
Explore more questions on app