# Homework Help: Horizontal ProjectileMotion

1. Oct 4, 2009

### cire792

[SOLVED] Horizontal ProjectileMotion

1. The problem statement, all variables and given/known data
Divers jump off a cliff, at a height of 35m above sealevel. To land safely, they aim for 5m from the base of the cliff. What push off speed is neccesary?

Sy= -35m
Sx= 5
ay= -9.8m/s/s
Vi = ?

2. Relevant equations
Not sure how to proceed, if Vfy = 0 at the end of the motion,
then vf^2 = vi^2 + 2a(s)

3. The attempt at a solution

Adds up to vi = 0, unless I really screwed up my math.

Last edited: Oct 4, 2009
2. Oct 4, 2009

### Staff: Mentor

Are you assuming that their initial velocity is purely horizontal?

Surely the final velocity (on hitting the water) is not zero!

Hint: How long does it take the diver to hit the water?

3. Oct 4, 2009

### cire792

Yes, does that then change initial velocity to 0?
Well that can't be right, at the point they jump off its 0,
so then I can just figure out the time with
Sy = vi(t) + 1/2 (a)(t)^2
Which equals 0.84s.
I don't understand how I'd continue with that.

4. Oct 4, 2009

### Staff: Mentor

Treat the horizontal and vertical components of motion separately. The vertical component of the initial velocity is zero. Use that to find the time of fall.
OK.
Show how you got that result.
Once you correctly figure out the travel time, you can use it to figure out the horizontal speed.

5. Oct 4, 2009

### cire792

Sy = vi(t) + 1/2 (a)(t)^2
-35 = 0(t) + (-4.9) (t)^2
t = 0.84s

And because x is uniform,
v=s/t
v=5/0.84
v=5.95m/s

So, that would be the X-component.
But since it's horizontal at the start,
Vi = 5.95m/s [E]
So problem solved.
Thanks a lot!

6. Oct 5, 2009

### Staff: Mentor

Redo this calculation.

7. Oct 5, 2009

### cire792

Ahh!
t = 2.6
Well that makes more sense.
Don't know how I messed that up before. ~_~
oh, vi = 13.46