# Projectile problem w/ unkown velocity/time

1. Dec 8, 2008

### bmyer

1. The problem statement, all variables and given/known data
An athlete executing a long jump leaves the ground at a 25° angle and lands 6.30 m away.
What was his take off speed?
2. Relevant equations
(In all equations, X can be substituted for Y in entire equation)
Vy= Voy + AyT
Vy^2 = Voy + 2AyY
Y=VoyT + (1/2)AyT^2

3. The attempt at a solution
The answer is 8.98 m/s. I can't figure out how to do this problem, and need to understand this for an upcoming final. I have worked several other similar problems, but none without a known velocity and/or time.
Any insight into how to even begin would be greatly appreciated.

Last edited: Dec 8, 2008
2. Dec 8, 2008

Sorry I edited this because I think i made a mistake don't worry I didn't give up on you.

I'm sorry again I keep getting an answer very close and is probably just off because of my rounding. Though it's been a while since I have done projectile motion and I would hate to show you the wrong way with a final coming up.

Last edited: Dec 8, 2008
3. Dec 8, 2008

### LowlyPion

The x component of velocity is Vo*Cosθ

His total distance would be given by Vo*Cosθ*t

So ...

X = 6.3m = Vo*Cos25*t

Now you still need time so ... look to the skies.

Vo*Sinθ = g*t and t to max height = Vo*Sinθ/g

But total time = 2*t

Now we can solve. t = 2*Vo*Sin25/g = 6.3/Vo*Cos25

Rearranging Vo2*2*Sin25*Cos25 = 6.3*g

Recognizing that 2*Sinθ*Cosθ = Sin2θ

Vo2 = 6.3*g/Sin50