A catapult on a cliff launches a large round rock towards a ship on ocean below.

Question

A catapult on a cliff launches a large round rock towards a ship on ocean below. The rock leaves the catapult from a height H of 34.0m above sea level, directed at an angle(theta) above the horizontal with an unknown speed (vo). The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 148.0m. Assuming that air friction can be neglected, calculated the value of the angle(theta). Calculate the speed at which the rock is launched. To what height above sea level does the rock rise?

Explanation / Answer

dy = u*t + 1/2 *g*t^2

-34 = vo*sin(theta)*6 + 1/2 * 9.8 * 6^2

vo*sintheta = 35.067 m

for horizontal displacement

dx = (v+u)*t/2

148 = [(vo*costheta + vo*costheta)(6)]/2

vocostheta = 24.67 m/s

vosintheta/vocostheta = 35.067/24.67

theta = tan^-1(35.067/24.67)

theta = 54.87 degree

part b )

dy = ut + 1/2 * gt^2

-34 = vo*sin54.87 (6) - 1/2 * 9.8 * 6^2

vo = 29.02 m.s

part c )

hmax = vo^2 sin^2(theat)/2g

hmax = 28.74

h = 34 + 28.74 = 62.74 m

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