# Snowball off of a roof

1. Oct 1, 2007

### ptpatil

[SOLVED] Snowball off of a roof

1. The problem statement, all variables and given/known data
A snowball rolls off of a roof angled downwards at 40.0 degrees, it has an initial velocity of 7.00 m/s. The roof is 14.0m tall.

Find how far from the edge of the roof the snowball falls

2. Relevant equations
I know that cos -40 x 7.00m/s is the horizontal velocity, and Vy initial is sin -40 x 7.00.

3. The attempt at a solution

OK, this seemed like a ridiculously easy problem, but the thing is I keep getting a value of t that i just know isnt correct (my friend already solved it, but I'd rather understand what I'm doing wrong than just copy).

To find t (time), I used Y= Yo + Vy x t + 1/2gt^2
I put in the values and had the ground as 0m and inital Y as 14.0m.

I got this: 0=14m + (-4.49m/s x t) + (-4.9 m/s/s x t^2)

I solved for t and always get 1.29s as the possible root. My friend says its incorrect and it seems incorrect to me as well since that would mean the snow ball lands a hefty 7 meters away from the roof.

Help?

2. Oct 1, 2007

### cristo

Staff Emeritus

As a check for your comment about the ball landing a hefty 7m from the roof, you can calculate the maximum distance it could land if there were no acceleration using trigonometry, so I wouldn't say that 7m seems unreasonable. However, for the reason I pointed out above, this is incorrect anyway.

3. Oct 1, 2007

### ptpatil

I've checked and rechecked my trig, i always come up with the same initial velocity of -4.49 m/s. =(

4. Oct 1, 2007

### cristo

Staff Emeritus
Ok, there's some ambiguity with the way you say "angled downwards." Well, at least in my mind anyway. However, from re-reading I guess you mean that the angle between the horizontal and the slope of the roof is 40 degrees? If so, your components are correct; sorry!

So, how did you solve the equation? Did you use the quadratic formula? If so, perhaps you could post your work and we can check for errors.

5. Oct 1, 2007

### ptpatil

Oh, sorry about that, I should've cleared that bit.

Anyways, I did use the quadratic formula,

i said: 4.49 +- sqrt( -4.49^2 - 4(-4.9)(14.0))/-9.8

simplified, i got: 4.49 +- 17.16 / -9.8, I took the 4.49-17.16/-9.8 and got 1.29 seconds.

I even checked with an online quadratic solver, and it gave me the same figs

6. Oct 1, 2007

### cristo

Staff Emeritus

7. Oct 1, 2007

### ptpatil

Well its good to hear that im atleast not delusional, I really dont know why im so sure im the one who's wrong, maybe he is, he would tell me that he got .89 seconds and 4.7 or something as the distance.

I don't know how he got it, but I think im going to keep my answer, thanks for the help.

8. Oct 2, 2007

### cristo

Staff Emeritus
You're welcome!