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. A) The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 160.0 m. 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?

Explanation / Answer

  for x direction:

Vx = D / t = 160 / 6 = 26.66

so Vo cos q = 26.66... (1)

for the vertical y motion :

- 31 = Vyt - 1/5 g t^2

=> -31 = Vo sin q (6) - 4.9 (36)

=> Vo sin q = 24.23 ... (2)

div (2) / (1) u get :

tan q = 0.8 => q = 38.66 degrees

- taking eqn 1 u get:

Vo cos 38.66 = 30.17

=> Vo = 38.6 m/s

- at max h : Vy = 0 , so:

0 = [38.6 sin(38.66)]^2 - 2(9.8) y

=> y = 29.7 m

- Vy = 0 at max height

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