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 32.0m 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 183m.

a.) The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 183m. Assuming that air friction can be neglected, calculate the value of the angle .

b.) Calculate the speed at which the rock is launched.

c.)To what height above sea level does the rock rise? (Whats the max height it reaches?)

Explanation / Answer

yi - yf = v0 sin(theta) t - g t^2 /2

0 - 32 = 6 v0 sin(theta) - 176.4

v0 sin(theta) = 24.07 ..... (i)

x = v0 cos(theta) t

v0 cos(theta) = 183 / 6 = 30.5 . .....(ii)

(A) (i)/(ii) => tan(theta) = 0.789

theta = 38.3 deg

(B) v0 = 39 m/s

(C) 0^2 - (39 sin38.3)^2 = 2(-9.8)h

h = 29.6 m

H_max = h + 32 = 61.6 m

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