A stone thrown horizontally from a height of 5.72m hits the ground at a distance

Question

A stone thrown horizontally from a height of 5.72m hits the ground at a distance of 14.50m. Calculate the initial speed of the stone. Neglect air resistance. 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 34.0m above sea level, directed at an angle above the horizontal with an unknown speed v_0. The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 158m. Assuming that air friction can be neglected, calculate the value of the angle (in degrees).

Explanation / Answer

1)

Drop time from 5.72m. = sqrt(2h/g) = sqrt(2*5.72/9.8) = 1.0804 secs.

Vo = 14.50 m / 1.0804 sec = 13.42 m/s

2)

Lets assume the angle is (X), and initial velocity is v0. Resolve the velocity into vertical and horizontal velocity.

Vertical = v0 sin X
Horizontal = v0 cos X
Total time = 6 s

The horizontal velocity is always constant. So,

(v0 cos X)(t) = 158 --------------(1)

It says that the rock hits a ship on the ocean BELOW. That means the displacement would be 34 m below.

Use the formula,

s = ut + 1/2at^2
34 = (vosin X)(t) + 1/2(-9.8)(6^2)
(v0 sin X)(t) = 210.4 ----------(2)

Take (2)/(1),

tan X = 1.3316
X = 53.1 degrees

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