Determine the vertical components of the launch velocity

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blrdey
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I have a problem that I can't figure out what formula to use. It states:

In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0588 s, during which time it experiences an acceleration of 290 m/s2. The ball is launched at an angle of 48.5 ° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.

Anyone know how exactly I'm supposed to do this?
 
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Use the acceleration to calculate the final velocity of the ball, that is its velocity when it loses contact with the boot.
 
Would that be V(final) = V(initial) + at ?
If so:
V(final) = 0 + (290m/s^2)(0.0588s)
V(final) = 17.052

I thought that the acceleration for the X-axis always has to be set at zero.
 
Yes. That is the launching velociy, vo, of the ball. The acceleration along the x-axis for a projectile is zero as you remarked, but while the ball is in contact with the boot it is not considered to be a projectile, only when it is free falling...
 
So far you're correct. When the ball leaves his foot it is traveling at 17.052 m/s.

Now gravity has its chance to work its magic. It will accelerate the ball towards the ground in the y-direction only. It will not affect the x-axis.

However, your question asks nothing about the fate of the ball after it leaves the kicker's foot. It only wants the x-component and y-component of a ball traveling 17.052 m/s at an angle of 48.5 degrees above the horizontal (the ground).

Dust off your trig skills.

Hint. 48.5 degrees is very close to 45 degrees. If he kicked it at 45 degrees, the x-component and the y-component would be equal. This is good practice for all physics problems. Off the top of your head come up with an approximate answer that you can compare your computed answer to. That way if you come up with an answer like (a) .00124 m/s, (b) 18743 m/s, you can conclude you did something wrong.
 
I can't figure out where to go from here.
 
Oh! Okay, thank you both!