How Far Does a Snowball Travel After Rolling Off a Roof?

[ptpatil]

[SOLVED] Snowball off of a roof

Homework Statement

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

Homework Equations

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

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 isn't 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?

 

[cristo]

I'd check your trigonometry and, more specifically, your vertical initial velocity.

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.

 

[ptpatil]

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

 

[cristo]

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.

 

[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

 

[cristo]

Well, your working looks correct to me, so I'd say your answer was too. What makes you so sure that your friend has got the correct answer?

 

[ptpatil]

Well its good to hear that I am atleast not delusional, I really don't know why I am so sure I am 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 I am going to keep my answer, thanks for the help.

 

[cristo]

You're welcome!