Can You Win the Puck Game at the Amusement Park?

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Homework Statement



One game at the amusement park has you push a puck up a long, frictionless ramp. You win a stuffed animal if the puck, at its highest point, comes to within 10 cm of the end of the ramp without going off. You give the puck a push, releasing it with a speed of 5.0m/s when it is 8.5m from the end of the ramp. The puck's speed after traveling 3.0m is 4.0 m/s. Are you a winner?

Homework Equations



vf^2 = vi^2 +2ad
x(t) = 1/2at^2 + vt

The Attempt at a Solution



a = (vf^2-vi^2)/2d
a = (16-25)/6
a = -3/2 m/s/s

x(t) = -3/2t^2 + 5t m
v(t) = -3t + 5 m/s
v(t) = 0 = -3t + 5
t = 5/3

x(5/3) = 26/5 m = 4.16m
df = 8.5 - 4.16
df = 4.3

Therefore, you are not a winner because you are off by a distance of 4.2m before you earn the stuffed animal at the amusement park.

Is this the correct method of doing this question. I feel that I am wrong because in high school it was always winning or going in the basket and not it is not. I believe it is unlikely that the answer will tell us that we are not a winner. Is there anything wrong with my logic process here?
 
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barthayn said:
a = (vf^2-vi^2)/2d
a = (16-25)/6
a = -3/2 m/s/s

This is correct.

barthayn said:
x(t) = -3/2t^2 + 5t m
v(t) = -3t + 5 m/s
v(t) = 0 = -3t + 5
t = 5/3
Check your relevant equations and your previous work. You have a slight arithmetical error here that makes a very large difference in your final solution.


Your process seems sound, just check your arithmetic. Post again if you want another double check.
 
Ignea_unda said:
This is correct.Check your relevant equations and your previous work. You have a slight arithmetical error here that makes a very large difference in your final solution.Your process seems sound, just check your arithmetic. Post again if you want another double check.

Thanks for pointing that out. I seen a two below the number and thought it was halved. It wasn't. I got it stopping 6.67cm away from the edge, and therefore I am the winner of the stuffed animal correct?