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 = 34.0 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 = 181 m.

A) Assuming that air friction can be neglected, calculate the value of the angle (in degrees).

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

C) What is the maximum height above sea level the rock will rise to?

Answer 1

velocity in x direction,

Vx = D/t = 181/6 = 30.17 m/s

Vx = v0 cos(thets) = 30.17 (1)

we know from the eqn of motion

S = ut + 1/2 a t^2

H = vo sin(theta) t - 1/2 g t^2

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

vo sin(theta) = 23.73 (2)

(2) / (1)

sin(theta)/cos(theta) = 23.73/30.17 = 0.7865

theta = tan-1 (0.7865) = 38.2 deg

Hence, theta = 38.2 deg ------------------Ans(A)

v0 sin(38.2) = 23.73

v0 = 38.37 m/s -------------------> Ans(B)

maximum height, we know from eqn of moiotn

v^2 = u^2 + 2 a S

0 = [ 38.37 sin(38.2)]^2 - 2 x 9.8 y

y = 28.73 m -----------------> Ans (C)