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 = 32.0 m above sea level, directed at an angle theta = 46.9° above the horizontal, and with a speed v = 29.6 m/s. Assuming that air friction can be neglected, calculate the horizontal distance D traveled by the projectile.

Answer 1

Vo = 29.6m/s[46.9o].
Xo = 29.6*Cos46.9. = m/s.
Yo = 29.6*sin46.9 = m/s.

Y = Yo + g*Tr.
Y = 0.
g = -9.8 m/s^2.
Tr = Rise time.
Tr = ?

Y^2 = Yo^2 + 2g*h.
Y = 0.
h = ?

0.5g*Tf^2 = h + 32.
g = 9.8 m/s^2.
Tf = Fall time.
Tf = ?

D = Xo*(Tr+Tf).