# Angle below horizontal projectile question

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1. Jul 12, 2015

### jdang

1. The problem statement, all variables and given/known data
• a ball rolls off an incline on top of a 9.0m building
• at a velocity of 22m/s
• 32° below the horizontal
• how far from the base of the building will the ball hit the ground?
• [ANS = 11m)
2. Relevant equations
• vf^2 = vi^2 + 2ad
• d = ((vf+vi)/2)t
• d = vt
3. The attempt at a solution
• x (um)
• v = 22cos32 = 18.657...m/s
• y (uam)
• a = -9.81m/s^2
• vi = -22sin32 = -11.658...m/s
• d = 9.0m

• vf^2 = vi^2 + 2ad
vf^2 = (-11.658...m/s)^2 + (2*-9.81m/s^2*-9.0m)
vf^2 = 312.494...m^2/s^2
vf = 17.677...m/s

• d = ((vf+vi)/2)t
d=((vf+vi)/2)t
-9.0m = ((17.677...m/s - 11.658...m/s)/2)t
t = ((17.677...m/s - 11.658...m/s)/2) / 9.0m
t = 0.334...s

• d = vt
d = (18.657...m/s)(0.334...s)
d = 6.495...m
d = 6.5m

I'm not sure where I made my mistake since I am about 4.5m off.

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Last edited: Jul 12, 2015
2. Jul 12, 2015

### Nathanael

Vi.y is positive (same direction as gravity, which you made positive)

Or if you made gravity negative, then Vf.y should be negative.

Either way it needs to be consistent.

3. Jul 12, 2015

### TSny

When taking a square root, you have two choices of sign. Which sign is appropriate here?
You didn't solve for t correctly here.

4. Jul 12, 2015

### jdang

Sorry, I calculated it negative but I forgot to type it up with the negative sign. I will make vfy negative, thank you!

5. Jul 12, 2015

Thank you!