# Projectile question -- ball rolling on a chair

1. Sep 19, 2014

1. The problem statement, all variables and given/known data

Hi all, in my grade 11 book there is a question, and the question is :- there is a ball rolling on a chair by a speed of 5.0 m/s on the X axis, the ball hit the ground on a distance of 0.25 m on the X axis, What is the Δy and what is the speed on Y axis

ϑx (speed on X axis) = 5.0 m/s

Δx = 0.25 m

Δy = ??

ϑy = ??

i think also i need the time, i tried to extract the time using the three equations that i will write them down.

t=??

2. Relevant equations

αy = -g = -9.81 m/s²

ϑfy = ay x t

ϑfy^2 = 2ay x Δy

Δy = 1/2ay x t^2

3. The attempt at a solution

Since all equations require time except ϑfy^2, I tried to find time first because i think it's the core of this question even if the question didn't need it, this is my attempt :-

http://im77.gulfup.com/kllied.jpg [Broken] << Huge image replaced with URL by Moderator >>

i don't know if the time (0.5 seconds) is correct or not, i need help please and i don't want the solution of the question i just need to know if the way i get the time is correct or not, and if it's not please help me getting the right way to solve this question, and thanks so much

Last edited by a moderator: May 6, 2017
2. Sep 19, 2014

### BvU

Hello Ado, and welcome to PF.
Big picture!
I see something going wrong at the top under "equations?"

$\Delta y = {1\over 2} g t^2$ and $v_y = gt$ can not be combined to $v_y = 2 g \Delta y$ !
Always check the dimensions when you make a step

At best you get something like $v_y^2 = 2 g \Delta y$ which you will later on recognize as energy conservation when potential energy (from height) is converted to kinetic energy (from speed).

And it doesn't help you to find t because both y and vy are unknown.

And you do surprise me when you write (at the very top) $v_x = 5$ m/s, $\Delta x= 0.25$ m, and then still follow with $t = ??$. Think again!

Finally, once you find your t = 0.5 s, don't you hear the alarm bells go off when you reconsider the 5 m/s and the 0.25 m ?

3. Sep 19, 2014

### nasu

You are right that you need to find the time.
However the equation you are trying to use is wrong.
You have a speed on the left hand side and a speed squared on the right hand side.
The equation you have in mind is possibly
v_f^2=2ay but you forgot the square.
If you write correctly the time (and everything else) will simplify, as expected. These equations you are trying to combine are not independent.

Besides the equations for the y directions you have some for the x directions. Why not write these too? You know a lot about the x direction. Speed, distance.

4. Sep 19, 2014

### BvU

Now, unless you made a typing error in rendering the exercise, the chair height that comes out gives the impression it's made for toy trains.

What would be a reasonable distance (given the 5 m/s) to get a reasonable chair height, of say, 0.5 m?

And the other way around: what would be a reasonable speed, given the 0.25 m distance travelled in x ?

5. Sep 19, 2014

Thanks a lot, i really feel stupid now, i seriously feel stupid :< Lol

i really didn't though of x by speed because i though i will not use this simple equations anymore xD

anyway, i think every thing in physics is relative now, and thanks a lot, you saved my day :D

6. Sep 19, 2014

Actually, my physics teacher in school solved another question, it's somewhat relative but that question was about the projectiles with corners, he just replaced Vy with (ViSinθ) so i said why i don't just replace delta Y with 1/2gt^2, thanks a lot guys i really appreciate this fast help :)

7. Sep 19, 2014

the distance is the distance between the chair base and the ball landing area, i can take a picture of the question but it's in arabic :/, and i tried to translate it with my best english knowledge, maybe i missed some words but i think the numbers are right :3

8. Sep 19, 2014

the speed of the ball on the X axis is :- 5 m/s

the ball passed 0.25 m on the X axis

the chair height is delta Y

Edit:- i will solve this question and i will take picture of it and upload it to see if my answer is correct or not

9. Sep 19, 2014

I got a very weird answer.

t = distance\speed = 0.25\5 = 0.05 seconds

Δy = 1\2 g t^2 = 1\2(-9.81)x(0.05)^2 = 0.012m (the answer in the calculator is -0.012m but the distance isn't targeted variable)

V_fy = gt = (-9.81)x0.05 = -0.4905 m\s

it's very weird because the speed on X axis is 5 m\s and the answer isn't reasonable as you said :/.