- #1
red888
- 1
- 0
Hi I am currently doing physics a level and am in a spot of bother not knowing what to do next.
My data analysis is to project a ball bearing down a curved ramp and off a 0.952m desk. After obtaining the results of how far away the ball lands I am now attempting to comapre them with the theoretical results I get though I am stuck as to how I am supposed to work out that part.
P.E. = K.e
P.E. = mgh = K.E. = ½ mv²
mgh = ½ mv²
v=u+at
s=ut+1/2at²
v²=u²+2as
P.E. = mgh = K.E. = ½ mv²
mgh = ½ mv²
gh = ½ v²
2gh = v²
2 x 9.8 x 0.2 = v²
3.92 = v²
√3.92 = 1.98
By placing 0.2 as a height interval I have now worked out the velocity that the ball shall travel at though this is not that helpfull as I have no idea how I am supposed to work out how quickly the ball shall drop due to gravity.
I would really appreciate some help as my textbooks have proven not very good at explaining this and my coursework is due in tommorow.
thanks,
Homework Statement
My data analysis is to project a ball bearing down a curved ramp and off a 0.952m desk. After obtaining the results of how far away the ball lands I am now attempting to comapre them with the theoretical results I get though I am stuck as to how I am supposed to work out that part.
Homework Equations
P.E. = K.e
P.E. = mgh = K.E. = ½ mv²
mgh = ½ mv²
v=u+at
s=ut+1/2at²
v²=u²+2as
The Attempt at a Solution
P.E. = mgh = K.E. = ½ mv²
mgh = ½ mv²
gh = ½ v²
2gh = v²
2 x 9.8 x 0.2 = v²
3.92 = v²
√3.92 = 1.98
By placing 0.2 as a height interval I have now worked out the velocity that the ball shall travel at though this is not that helpfull as I have no idea how I am supposed to work out how quickly the ball shall drop due to gravity.
I would really appreciate some help as my textbooks have proven not very good at explaining this and my coursework is due in tommorow.
thanks,