# How to work out the drop of a ball once the velocity is known

Hi im currently doing physics a level and am in a spot of bother not knowing what to do next.

## 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 appriciate some help as my textbooks have proven not very good at explaining this and my coursework is due in tommorow.

thanks,

Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
Hi im currently doing physics a level and am in a spot of bother not knowing what to do next.

## 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 appriciate some help as my textbooks have proven not very good at explaining this and my coursework is due in tommorow.

thanks,
The v you have is the initial (horizontal) velocity when the ball comes off the curve and begins to fall. The equations you give, above, don't take x and y components into account. Fortunately, the components can be handled separately. How long would it take a ball, with no initial (downward) velocity, fall 0.952m? Since gravity does not affect the horizontal motion, how far will the ball move horizontally in that time?

Taking into account what has been said by HallsOfIvy, the following little ditty should be used:

s=ut+1/2at²

More information as to the shape of the ramp may be in order... Is the portion where the ball leaves the ramp horizontal i.e. parallel to the x-axis?

djeitnstine
Gold Member
Make sure to break your model down into components, for instance the ramp: I assume its a diagonal line and not a curve. And the desk I assume is a straight line. And as for how far away the ball will fall. Simply use your velocity obtained as it rolls along the desk and find the distance traveled as it falls off of the desk.

btw Brendan, your name has an 'a" in it and mine has an "o" pretty kool attempt to steal my first name ;) j/k