Question

A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H = 31.0 m above sea level, directed at an angle theta = 47.9° above the horizontal, and with a speed v = 30.9 m/s. Assuming that air friction can be neglected, calculate the horizontal distance D traveled by the projectile.

if the answer involves using quadratic formula to find time, please show all work. thank you!

Answer 1

initial vertical component (Voy) of velocity = 30.9 sin 47.9 = 22.92 m/s
initial horizontal component (Vox) of velocity = 30.9 cos 47.9 = 20.71 m/s

time to reach max height = t = Voy/g = 22.92 / 9.81 = 2.33 s
max height above cliff = h = Voy(t) - 1/2gt² = (22.92)(2.33) - (0.5)(9.81)(2.33)² = 26.80 m
max vertical height = 31.0 + 26.80 = 57.8 m

time to fall vertically 57.8 m:
H = 1/2gt²
t² = 2H/g = (2)(57.8)/9.81 = 11.78
t = 3.43 s

Total time of travel in air:
T = 2.33 + 3.43 = 5.76 s
Horizontal distance of rock travel = (20.71)(5.76) = 119.33 m