Projectile Motion: Angle of Impact

[stripedcat]

Homework Statement

A ball is thrown from a bridge 200 m high with an initial velocity of 34.4 m/s at an angle of 54° above the horizontal.

Homework Equations

I'm not sure... arctan Vx/Vy = x? I really don't know.

The Attempt at a Solution

(a) How high above the throwing height does the ball go: 39.5 meters
(b) The total time the ball is in the air: 9.83 seconds
(c) The horizontal distance the ball travels before it hits the water: 199 meters
(d) The velocity of the ball as it strikes the water: 71.4 meters/second

And here is the one I don't know how to calculate.

Direction _______ degrees below the horizontal.

I know the answer to be 73.6 degrees, I do not know how to calculate the answer. I was looking for an online resource, but I could only find information under the assumption of a level field as opposed to launching the projectile from a height. I was able to fairly easily calculate the other numbers there, but ran into an issue trying ti figure out the last part.

 

[Simon Bridge]

Welcome to PF;
You already know how to do this without another resource.
The trick is to split the velocity into horizontal and vertical components.
You can work out the horizontal and vertical velocities at the time of impact - sketch them out head-to-tail and find the angle the resultant makes to the horizontal. Yes - it's the arctan.

 

[stripedcat]

Oooh, had the wrong Vx/Vy, or the wrong Vy rather. That was the problem. Thanks.