# Formula for impact velocity at an angle

1. Oct 9, 2009

### laurmaso

1. The problem statement, all variables and given/known data
A projectile is launched from a cliff that is 84m tall with an initial speed of 32m/s at an angle of 25 degrees above the horizontal. The object is in the air for a total of 5.74s. What is its impact velocity and at what angle?

2. Relevant equations
Vfy = Voy + ayt

Vf^2 = Vfx^2 + Vfy ^2

tan(theta) = (Vfy)/(Vfx)

3. The attempt at a solution
I have that the answer is 51.7m/s at an angle of 55.9 degrees below the horizontal. However, when I attempt to solve, my answers don't match up:

Vfy = Voy + ayt
= 0 + (-9.8m/s^2)(5.74s)
= -56.25

Vf^2 = Vfx^2 + Vfy^2
= (32)^2 + (-56.25)^2
= 1024 + 3164.06
= 4188.29
Vf = 64.71m/s <------ incorrect

Then when I plug in to find the angle of impact:

tan(theta) = (Vfy)/(Vfx)
= (-56.25)/(32)
= -1.7578
tan^-1 (-1.5758) = -60.36 degrees below the horizontal <------- also incorrect.

I'm not sure what I'm doing wrong. Any suggestions??

I also need to calculate the peak height of the projectile; I first used:
t = (Voy)/g
= 32sin25/(-9.8m/s^2)
= 13.5/(-9.8)
= 1.37s

then y = (1/2)at^2 + Voyt
= (-4.9m/s^2)(1.37s)^2 + 13.5m/s(1.37s)
= -9.2 + 18.5
= 9.3m <-------- this answer cannot be correct, not sure what I'm doing wrong

Last edited: Oct 9, 2009
2. Oct 10, 2009

### tiny-tim

Welcome to PF!

Hi laurmaso! Welcome to PF!
Sorry, but this makes no sense

Voy isn't 0, and you haven't used the 84m.

Try again.