Angle below horizontal projectile question

[jdang]

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.

 

[Nathanael]

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

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

Either way it needs to be consistent.

 

[TSny]

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]

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

Or if you made gravity negative, then V_f.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!

 

[jdang]

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!