help with physics, projectile motion

[Littledude]

So I have a lab report to do and I'm working on the equations bit.
The lab was basically shooting a ball out of a projectile launcher (this kind to be exact) and recording the distance it travels.
For the first set of trials we set the angle of the launcher to 0 and recorded that data,  my lab partners calculated the initial velocity of the ball to be 8.9 m/s for this.
Then we set the launcher to an angle of our choosing which was 26. Before we started the trials we were supposed to predict the distance the ball would travel, we didn't do this and now I'm stuck doing it alone, but I have no idea how.
The data I have to work with is the initial velocity: 8.9 at 26 degrees and the height of the launcher's barrel above the ground which is .26 meters.
how can I find the distance the ball would travel, neglecting air resistance with the above parameters?

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[matthewbim11]

Something about inertia, I don't know.
http://www.physicsforums.com/showthread.php?t=261743

[Littledude]

hmm
I think I figured out how to get time down
so in theory the only question i have is if the initial and final velocity in the horizontal direction would be equal when there is an acceleration of 0

[matthewbim11]

Yeah, I'm in eighth grade. Ask that website I gave you.

[Littledude]

ugh
The distance I got is more than double the measured distance

[Nexus]

This was the very first week of physics for me.

use the equation distance = initialvelocity*time + (1/2)*acceleration*(time)^2 with only the vertical component of the velocity to find the time it takes the ball to hit the ground

You can find the initial velocity by taking what velocity you found the horizontal launcher to launch at, and multiplying that by sin(26) to find the vertical component.

Once you find time, and you know that the horizontal component of the velocity is constant (multiple above by cos(26) instead of sin to get it) then you can use the equation v = d/t to find distance

Also post the distance that the horizontal ball went, and the height if it is different.

[OneWithFire]

if i remember correctly, this kind of thing uses 3 formulas.

v_f=v_i+at

v_f²=v_i²+2ad

d=v_it+.5at²

then again, we went over this the first 2 days in college physics, was all harder from then on.

[Littledude]

This was the very first week of physics for me.

use the equation distance = initialvelocity*time + (1/2)*acceleration*(time)^2 with only the vertical component of the velocity to find the time it takes the ball to hit the ground

You can find the initial velocity by taking what velocity you found the horizontal launcher to launch at, and multiplying that by sin(26) to find the vertical component.

Once you find time, and you know that the horizontal component of the velocity is constant (multiple above by cos(26) instead of sin to get it) then you can use the equation v = d/t to find distance

Also post the distance that the horizontal ball went, and the height if it is different.

the ball went down .26 m from the starting position and traveled 2.23 m

[Nexus]

t = sqrt(2d/a) = 0.23 seconds

v = d/t = 9.696 m/s

hmm, your lab parter probably got confused about which distances to use where.


Vy = 9.696*sin(26) = 4.25 m/s

-0.23 = 4.25*t + 0.5(-9.8)t^2

solve for t...
t = -0.511 or 0.918

negative time makes no sense, so use 0.918.

Vx = cos(26)*9.696 = 8.7147 m/s
d = v*t = 8.0001m

[Littledude]

t = sqrt(2d/a) = 0.23 seconds

v = d/t = 9.696 m/s

hmm

nah
it would be time = sqrt(-.26/(.5*-9.8)) = 0.23 seconds
and the measured distance for the first set of trials was 1.105
which would end up being 4.8?
so I guess they are still wrong?

[Nexus]

nah
it would be time = sqrt(-.26/(.5*-9.8)) = 0.23 seconds
and the measured distance for the first set of trials was 1.105
which would end up being 4.8?
so I guess they are still wrong?

what is 1.105 from?  did you give me the wrong distance?

I used 2.23

[Littledude]

what is 1.105 from?  did you give me the wrong distance?

I used 2.23

well there was one set of trials which were meant to be used to determine the initial velocity of the barrel, this was done at a 0 degree angle. the distance for those trials was 1.105.

but regardless I believe I figured it out, I just managed to get a distance value of 2.28 only sbout .05 off the measured so I'm good. the main problem was I had the wrong initial velocity
thanks for the help