NEETPhysicsMotion in Two Dimensions
A body of mass m is projected from the ground with an initial kinetic energy K at an angle of 30^ with the vertical. The kinetic energy of the body at the highest point of its trajectory will be
Options
- A3K 4
- BK 2
- CZero
- DK 4
Correct answer
D. K 4
Step-by-step solution
Let the initial speed of the body be u . The initial kinetic energy is given by: K = 1 2 mu^2 The angle of projection with the vertical is 30^ . Therefore, the angle of projection with the horizontal is = 90^ - 30^ = 60^ . At the highest point of the trajectory, the vertical component of velocity is zero. The speed of the body is equal to the horizontal component of its initial velocity, which remains constant. Speed at highest point, v = u 60^ = u 1 2 = u 2 The kinetic energy at the highest point is: K' = 1 2 mv^2