A projectile is launched with a speed of 41 m/s at an angle of 55° above the horizontal. Use conservation of energy to find the maximum height reached by the projectile during its flight.

Answer 1

Maximum height is achieved when the vertical velocity component, V sin 55 = 0

The kinetic energy loss at that time, which is due to the vertical component only, equals the potential energy gain.

(M/2) (Vsin55)^2 = M g H

H = (Vsin55)^2/(2g)