Physics 101 lab help should be easy

[dirtbird]

Okay, I'm not sure how I'm going to do this, but let's try. it is my third week of physics 101, and everything makes perfect sense, or so i thought. Our second lab, we used a ball shooter, and we shot one ball from a table onto the floor, one ball straight into the air, and one ball at a 10 degree angle. Guess which one I'm having problems with!
anyway, we shot the ball from a 10 cm height, at 10degrees, and the ball landed 199 cm away from the point of origin. So i set the problem up like this:
Yfinal= -.1 m
Yinitial= 0 m
Vyinitial= unknown
Vyfinal= unknown
A=-9.8m/s sq
T= unknown

Xfinal= 1.99 m
Xinitial= 0 m
Vx= unknown
A= 0 m/s sq
T= unknown

I have tried to explain how i tried to solve this with text, but i think i am not advanced enough to do that! essentially, i did some messy stuff with sin and cos, and came out with an initial velocity of 7.44 m/s. this is way off from the 5.75 i got for the first two experiments, and my lab partners got 2 completely different answers. I was wondering if it would be possible for someone to try to help me set up the problem. I can post what i did before if i need to, but it is proving very difficult! Thanks for any help
charla

 

[Suicidal]

I may not be the best person to help you. I just started to learn physics on my own, over summer break.
Here goes nothing:
I think you use the basic equation for projectile motion. (ignoring air resistance)
x = v*cos(10)*t
y=v*sin(10)*t – (1/2)*g*t^2

v is the initial velocity and g is acceleration due to gravity (g=9.8 m/s^2)

setting x = 1.99m and solving for t we get the time of impact

t= 1.99/( v*cos(10))
setting y = -.10 m in the second equation and substituting for t:
-.10 = v*sin(10)*(1.99/( v*cos10))– (1/2)*g*(1.99/( v*cos10))^2
.10 + tan(10)*1.99 = (1/2)*g*(1.99/( v*cos10))^2
v^2=[(1/2)*g*(1.99/( v*cos10))^2] / [.10 + tan(10)*1.99]
v=6.66 m/s

Hope this helps you.

Someone should check my work I may have made some careless mistake.

 

[quantumdude]

I see that Suicidal inserted numbers into the relations before he did the algebra. In case that is not clear to you, let me do it without doing the substitutions so you can see why the relation is as it is.

You have:

(1)...x(t)-x₀=v₀cos(θ)t
(2)...y(t)-y₀=v₀sin(θ)t-(1/2)gt²

Now, you didn't measure the time, but that's OK because we can eliminate the time to get y as a function of x.

From eq. (1):

t=(x-x₀)/(v₀cos(θ))

Insert that into (2) to get:

y-y₀=v₀sin(θ)(x-x₀)/(v₀cos(θ))-(1/2)g((x-x₀)/(v₀cos(θ))²

Since you used x₀=y₀=0m, we can drop those. Doing a little simplification yields:

y(x)=(v₀tan(θ))x-(g/(2cos(θ)))x²

which is a downward-opening parabola, as expected. That equation should also be in your book.

Here's a nifty site called HyperPhysics for more:

http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra2

 

[dirtbird]

Awesome, thanks a lot you guys! What i did it turns out was right, but it was much more complicated than what you guys did, so i think i made a math mistake somewhere in it. My professor said she would give me full credit for what i did, but i am personally not satisfied. . I'm going to work on it some more with the formulas you guys gave me, and i'll bookmark that awesome website, so hopefully i won't have to ask any more lame questions :P. Thanks again!
charla