# How Do You Find Launch Speed?

1. Sep 7, 2011

### kman2027

A cart launches a ball 15cm high. The vertical distance from the lowest ball to the highest is about two cart heights or 30cm.

Estimate the launch speed of the ball?

Estimate the time interval between successive stroboscopic exposures?

I honestly have no idea where to begin with either problem even after thoroughly reading the chapter where it is supposedly covered.

Does the equation y=h-(1/2)(g)(x/v0)^2 have anything to do with the problem?

2. Sep 7, 2011

hi kman2027!
??

3. Sep 7, 2011

### kman2027

That is all the text that is in the question, but there is a picture.

If you google image "cart launching a ball," the first picture that comes up is it.

4. Sep 7, 2011

### tiny-tim

ok, i see now
(i assume it means that the maximum height the ball reaches is 30 cm)

you have distance, acceleration, and final speed, and you want to find initial speed …

so which of the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations should you use?

Last edited by a moderator: Apr 26, 2017
5. Sep 7, 2011

### kman2027

Ok, I figured out the first part. The answer is 2.4m/s. The second part has me confused. I believe the question is asking how long the ball is suspended in air from the time the cart launches it till the time the cart catches it, but i'm not sure.

Estimate the time interval between successive stroboscopic exposures?

I've tried using the following two formulas for time and they don't seem to work:
t=-h/voy
t=sqrt(2h/g)

Any thoughts?

6. Sep 8, 2011

### tiny-tim

Yup!

(but it would be better f you showed your calculations)
pleeease stop using these formulas

the first only works for zero acceleration

the second only works in special cases

you need to learn the three standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations, and always to start with one of them

in this case, you have s = 0, v0 = 2.4, and a = -9.81, and you want to find t​

Last edited by a moderator: Apr 26, 2017
7. Sep 8, 2011

### kman2027

I did:
0=.3+2.4t+1/2(-9.8)t^2

t=.59

Is that the final answer or do you multiply t by 2?

8. Sep 8, 2011

not correct

9. Sep 8, 2011

### kman2027

i think i got the initial height (yo) and distance traveled (y) mixed up
does this look any better?
.3=0+2.4t+1/2(-9.8)t^2

-4.9t^2+2.4t-.3=0

t=.28???

10. Sep 8, 2011

### tiny-tim

that's better!

(and then double it to get the time between launch and return)

though slightly quicker would have been to use t = (v0 - v1)/a

(and i'm off to bed :zzz:)

11. Sep 8, 2011