King arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 12 m above the moat. the rock is launched at a speed of 30 m/s and an angle of 28 ∘ above the horizontal. how far from the castle wall does the launched rock hit the ground?

First find the vertical and horizontal components of the initial launch velocity. Vertical will be 30sin(28) and horizontal will be 30cos(28).

Next, find out how long it will be in the air using a form of the equation
X = Xo + Vot + (at^2)/2 (using only the vertical component of the initial launch velocity).

X is -12 m

-12 = 0 + 14.08*t + (-4.9)t^2

Solving for t we get 3.56 seconds.

Next, we multiply time by horizontal velocity to get horizontal distance.

3.56 * 30cos(28)

3.56 * 26.5 = 94.34m and that is the final answer.