- #1
AznBoi
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We are doing a Lab on Projectile motion:
http://online.cctt.org/physicslab/content/Phy1/labs/energy/Bullseye.asp
It is where you take a ball and roll it off a ramp, off the table. Then you would need to estimate the precise landing spot on the ground.
Directions: You are given the height of the table, but not the distance or the velocity of the ball rolling down. What you need to find is the distance (displacement of the ball from the table to the ground).
I have looked over many problems like this and I have solved examples many times and were correct. However, how do you find the velocity of the ball? My teacher said you don't really need the velocity to calcuate the distance but how??
Is this how you calculate the horizontal velocity of the ball coming down a ramp?
PEtop = KEbase
mgh = ½mv^2
2mgh = mv^2
square root( 2gh) = v
So would the velocity equal the sqr root(2(-9.8m/s^2)(h))?
Does the average horizontal velocity always equal that no matter how big the mass of the ball/marble is?? Please explain. Thanks.
After I know the velocity, I will be able to solve the lab, I just need to know if that equation is correct or not, and why. We haven't gone over PE and KE stuff.. Thanks a lot!![Big Grin :biggrin: :biggrin:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
http://online.cctt.org/physicslab/content/Phy1/labs/energy/Bullseye.asp
It is where you take a ball and roll it off a ramp, off the table. Then you would need to estimate the precise landing spot on the ground.
Directions: You are given the height of the table, but not the distance or the velocity of the ball rolling down. What you need to find is the distance (displacement of the ball from the table to the ground).
I have looked over many problems like this and I have solved examples many times and were correct. However, how do you find the velocity of the ball? My teacher said you don't really need the velocity to calcuate the distance but how??
Is this how you calculate the horizontal velocity of the ball coming down a ramp?
PEtop = KEbase
mgh = ½mv^2
2mgh = mv^2
square root( 2gh) = v
So would the velocity equal the sqr root(2(-9.8m/s^2)(h))?
Does the average horizontal velocity always equal that no matter how big the mass of the ball/marble is?? Please explain. Thanks.
After I know the velocity, I will be able to solve the lab, I just need to know if that equation is correct or not, and why. We haven't gone over PE and KE stuff.. Thanks a lot!