1. Sep 23, 2007

### jbutl3r25

1. A ball is thrown horizontally from the top of a building 22.8 m. high. The ball strikes the ground at a point 52.1 m. from the base of the building.
a. Find the time the ball is in motion.
b. Find the initial velocity of the ball.
c. Find the x component of its velocity just before it strikes the ground.
d. Find the y component of its velocity just before it strikes the ground.

2. vx =
dx=52.1
t

vfY
voY = 0
aY = -9.8 (gravity)
dY= -22.8
t

3. I found the time to be 2.16 s., but the rest is tricking me up. The answer to b can't be zero can it? I don't understand that part, and i'm just at a loss for how to find the x and y components. So please...can anybody give me some help here?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 23, 2007

### mgb_phys

You found the time from the vertical free fall?
There are no horizontal forces acting on the ball so the horizontal velocity must be constant. You know how far it went and how long it took so you have horizontal velocity.

You can find the vertical velocity from v^2 + u^2 + 2as, you know the initial vertical velocity is zero because it was thrown horizontally.

3. Sep 23, 2007

### jbutl3r25

I found the time from d=vo(t) + 1/2 at^2
And never mind about the horizontal, because i found that with dx=vx(t)
The part thats still tricking me up is the vertival...I've never seen v^2 + u^2 + 2as before so idk how else i would find it

4. Sep 23, 2007

### mgb_phys

Alternatively the vertical speed is v = u + a t, or simply v = gt if it is just dropped, this is obvious from the definition of accelaration.
the v^2 = u^2 + 2 a s, is from substiting the t from "s = ut + 1/2at^2" in "v = u + at"

5. Sep 23, 2007

### jbutl3r25

ohhhhh i think im getting that now...but then how would you find the initial velocity? it sounds like the easiest part but i still don't know how to fund it.