Question

A shot is fired at an angle of 60 degrees to the horizontal with kinetic energy E. If air resistance is ignore, what is the kinetic energy at the top of the trajectory?
a) zero
b) E/8
c) E/4
d) E/2

Answer 1

At the top of the trajectory, \(v_y=0\) so you only need to consider \(v_x=v\cos(\theta)\)

Answer 2

So \[
E_2 = \frac{1}{2}mv_x^2 = \frac{1}{2}m(v\cos(60^\circ))^2 = \left(\frac{1}{2}mv^2\right)\cos^2(60^\circ)=E\cos^2(60^\circ)
\]

Answer 3

It would be E/4. Thank you! :)