What Push Off Speed is Necessary for Divers to Safely Land 5m from a Cliff?

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[SOLVED] Horizontal ProjectileMotion

Homework Statement


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 = ?

Homework Equations


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

The Attempt at a Solution



Adds up to vi = 0, unless I really screwed up my math.
 
Last edited:
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Discussion

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?
 
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.
 
cire792 said:
Yes, does that then change initial velocity to 0?
Well that can't be right, at the point they jump off its 0,
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.
so then I can just figure out the time with
Sy = vi(t) + 1/2 (a)(t)^2
OK.
Which equals 0.84s.
Show how you got that result.
I don't understand how I'd continue with that.
Once you correctly figure out the travel time, you can use it to figure out the horizontal speed.
 
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!
 
cire792 said:
Sy = vi(t) + 1/2 (a)(t)^2
-35 = 0(t) + (-4.9) (t)^2
t = 0.84s
Redo this calculation.
 
Ahh!
t = 2.6
Well that makes more sense.
Don't know how I messed that up before. ~_~
oh, vi = 13.46