A shot is fired at 30° with the vertical from a point on the ground with kinetic energy K….

Q: A shot is fired at 30° with the vertical from a point on the ground with kinetic energy K. If air resistance is ignored, the kinetic energy at the top of the trajectory is

(a) 3K/4

(b) K/2

(c) K

(d) K/4

Ans: (d)
Angle with vertical = 30°

Hence Angle with horizontal = 60°

Initial K.E is

$ \displaystyle K = \frac{1}{2}m u^2 $

Velocity at the top is

v = u cos60 = u/2

K.E at top is
$ \displaystyle K’ = \frac{1}{2}m v^2 $

$ \displaystyle = \frac{1}{2}m (\frac{u}{2})^2$

= K/4