A catapult on a cliff launches a large round rock towards a ship on the ocean be

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 of 33m above sea level, directed at an angle theta above the horizontal with an unknown speed Vo.

The projectile remains in flight for 6 seconds and travels a horizontal distance D of 155m. Assuming that air friction can be neglected, calculate the value of the angle theta.

Calculate the speed at which the rock was launched.

To what height above sea level does the rock rise.

Explanation / Answer

a)

From the given we can write that
y(6)=0=33+v0*sin(%u03B8)*6-.5*g*36
and
x(6)=155=v0*cos(%u03B8)*6

rearrange the equations

18*g-33=v0*sin(%u03B8)*6

155=v0*cos(%u03B8)*6

divide
tan(%u03B8)=(18*g-33)/155

42.7738051221degrees above the horizontal

b)

speed of rock = 155(6cos42.77) = 35.1933390162 m/s

c)

v = v0 sin%u03B8-g*t=0

t = v0sin%u03B8/g =

vertical distance travelled

y = 33m+v0sin%u03B8*v0sin%u03B8/g-1/2*g*(v0sin%u03B8/g)^2

= 33+v0^2sin^2%u03B8/2g

by substituting values we get

y = 96.0009088458 m

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