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 31.0 m above sea level, directed at an angle above the horizontal with an unknown speed v0.

The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 180.0 m. Assuming that air friction can be neglected, calculate the value of the angle .

Calculate the speed at which the rock is launched.

To what height above sea level does the rock rise?

Explanation / Answer

a)

Vertical Direction
y = -31
a = -9.8
t = 6
y = vt + 0.5at^2
solving, v = y/t - 0.5at
plugging in numbers, v = 24.2333 m/s

Horizontal Direction
x = 180
t = 6
v = x/t = 30 m/s

(theta) = tan^(-1) [vy/vx]
(theta) = tan^(-1) [24.23/30]
(theta) = 38.93 degrees

b)

Vertical Direction
v1 = 24.23
v2 = 0
a = -9.8
h = [(v2)^2 - (v1)^2]/(2a)
h = 29.96 m
h + H = 60.1 m

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