# Angle below horizontal projectile question

## Homework Statement

• 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)

## Homework Equations

• vf^2 = vi^2 + 2ad
• d = ((vf+vi)/2)t
• d = vt

## 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:

Nathanael
Homework Helper
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.

jdang
TSny
Homework Helper
Gold Member
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
When taking a square root, you have two choices of sign. Which sign is appropriate here?
-9.0m = ((17.677...m/s - 11.658...m/s)/2)t
t = ((17.677...m/s - 11.658...m/s)/2) / 9.0m

You didn't solve for t correctly here.

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

When taking a square root, you have two choices of sign. Which sign is appropriate here?

You didn't solve for t correctly here.
Thank you!