Question

a 0.50kg projectile leaves the group with a kinetic energy 220j. at the heighest point of its trajectory, its kinetic energy is 120j. to what vertical height, relative to its starting point, does it rise?

Answer 1

1/2 M V^2 = 220 initial KE

V = (440 / .50)^1/2 = 29.7 m/s initial speed

1/2 M Vx^2 = 120 horizontal KE is constant

Vx = (240 / .5)^1/2 = 21.9 m/s horizontal speed

Vy = (29.7^2 - 21.9^2)^1/2 = 20.1 m/s initial vertical speed

t = 20.1 / 9.80 = 2.05 sec time to rise to max height

H = 1/2 g t^2 = 1/2 * 9.80 * 2.05^2 time to fall from height H

H = 20.6 m

Check:

H = Vy t - 1/2 g t^2 = 20.1 * 2.05 - 1/2 * 9.8 * 2.05^2 = 20.6