Variable Problem for 2D Kinematics

[apoptosis]

Here's another variable problem

Homework Statement

An extreme sportsman jumps from one cliff to another of width (w) by leaping horizontally by vo. With a height difference of h (where the starting cliff is higher) what is vo so he doesn't miss the cliff? (also, what is the direction of motion)

Homework Equations

xo=0
x=w, y=0
vox=0
ax=0

voy=0
ay=-g
yo=h

The Attempt at a Solution

Determine time of landing:
0=y=h-(1/2)gt$$^{2}$$
t=$$\sqrt{2h/g}$$

Determine vo:
x=vot
vo=xo/t=x$$\sqrt{g/2h}$$
vo=w$$\sqrt{9.81/2h}$$

does this look correct so far?

I'm stuck at finding the direction of motion, i know the equation is:
$$\vartheta$$=tan$$^{-1}$$vy/vx
vy=-gt=$$\sqrt{2hg}$$

=tan$$^{-1}$$$$\sqrt{2hg}$$/(w$$\sqrt{g/2h}$$)
=tan$$^{-1}$$ (2h/w) degrees of the horizontal component

but, I'm not sure if this is exactly right...

 

[Shooting Star]

The Attempt at a Solution

Determine time of landing:
0=y=h-(1/2)gt$$^{2}$$
t=$$\sqrt{2h/g}$$

OK.

How much time does he take to cross a horz dist of w, if he starts off with a horz velo of v0? Equate that time with the time you have got.

The final dircn is given by tan(theta) = vy/vx, at that point.

 

[apoptosis]

Thank you for your response.
I was wondering if you could please elaborate on your post.
My work is in my original post where i substituted time into the vot equation. Is this what you meant by equating time with the time that i have? (not quite sure what you mean)

also, i have used that tan ratio for direction of motion. does my resultl look reasonable? or is there a way i can simplify the variables?

 

[JolleJ]

The time to make the jump: Solve for t in: h = ½*g*t^2.
And then solve for v0 in: s = v0*t.

For the direction, just make the two values an vector, and determine the angle.

 

[Shooting Star]

Determine vo:
x=vot
vo=xo/t=x$$\sqrt{g/2h}$$
vo=w$$\sqrt{9.81/2h}$$

does this look correct so far?

It is absolutely correct. Somehow, I overloked this part. Sorry for that.

I'm stuck at finding the direction of motion, i know the equation is:
$$\vartheta$$=tan$$^{-1}$$vy/vx
vy=-gt=$$\sqrt{2hg}$$

Where's the minus sign of (-gt) gone? The value is OK. The direction is given by tan(theta) = -2h/w, where theta is the angle the velo makes with the x-axis, that is, the direction of v0. Since the body is moving downward and to the right, the angle is negative.