Solve King Arthur's Rock Launching Problem

[Sylis]

Homework Statement

i
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 25m/s and an angle of 30° above the horizontal. How far from the castle wall does the launched rock hit the ground?

Homework Equations

t_i=0, x_i=0, v_i=25, v_f=0, a=-9.8
Vf²=Vi² +2aΔx

The Attempt at a Solution

So I have a picture drawn of a horizontal line representing the ground/moat, and at one end a vertical line which extends 12m to signify the wall, and then another horizontal line signifying the "above the horizontal" part of the problem, and then a line at a negative slop that is at a 30 degree angle which also connects with the wall/horizontal. I believe I'm visualizing this correctly.

I get that horizontal velocity would be 25cos30. I'm just not quite sure what to do with that information. I sat here and thought about it for a bit and figured that if I used t_i=0, x_i=0, v_i=25, v_f=0, a=-9.8, I could use Vf²=Vi² +2aΔx to find the max height of the rock, and solve it in pieces that way, but that doesn't seem like the best use of the information I'm given, and also sort of defeats the purpose of using angles which is part of the lesson. Any thoughts?

 

[PhysicsandSuch]

Solving it in pieces is actually a great way to work these problems. You seem to be on the right track--maybe the angles lesson is only testing your ablility to properly decompose the velocity, and use each piece appropriately. Treat projectiles as one part vertical motion, and one part horizontal independently.

Once you find your maximum height (and note, your $v_i≠25$ here for finding that out->work out the vertical projection for this like you did the horizontal), think about using that information to find out how long the projectile stays in the air. From there, you should be able to work out the distance using the horizontal velocity you already discovered.

 

[CWatters]

One thing you know is that the flight time is fixed. Whatever flight time results from solving the vertical equations can be used in the horizontal equations.