Posted by Anonymous on Friday, February 8, 2013 at 8:51am.
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 = 35.0 m above sea level, directed at an angle theta = 45.9° above the horizontal, and with a speed v = 32.3 m/s. Assuming that air friction can be neglected, calculate the horizontal distance D traveled by the projectile.

Physics  Henry, Saturday, February 9, 2013 at 4:27pm
Vo = 32.3m/ @ 45.9o.
Xo = 32.3*cos45.9 = 22.5 m/s.
Yo = 32.3*sin45.9 = 23.20 m/s.
Y = Yo + g*t.
Tr = (YYo)/g = (023.2)/9.8 = 2.37 s.
= Rise time.
Tf1 = Tr = 2.37 s. = Fall time from hmax
to 35 m above sea level.
h = Yo*t + 0.5g*t^2 = 35 m.
23.2*t + 4.9t^2 = 35
4.9t^2 + 23.2t  35 = 0.
Use Quadratic Formula and get:
Tf2 = 1.20 s. = Fall time from 35 m above sea level to sea level.
T = Tr + Tfi + Tf2.
T = 2.37 + 2.37 + 1.20 = 5.94 s. = Time
in air.
D = Xo * T = 22.5m/s * 5.94s = 133.7 m.