# Kinematics: Projectile Motion, Constant Acceleration

1. Sep 5, 2013

### SammyLP250

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

Determine the minimum initial velocity, v0, and corresponding angle, θ0 needed to kick the ball just over the 3m high fence.

x = 6m
y = 3m

2. Relevant equations

X:
vx=v0x
x = x0+v0xt
v0x = v0cosθ
Y:
vy2=viy2
y = y0+v0yt + .5gt2
vy2 = v0y2+ 2g(s-s0)
v0y = v0sinθ
3. The attempt at a solution

I first started solving for v0y by setting the vy equal to zero and the y = 3 in the third Y equation.

0 = v0y2 +2(-9.81)(3)
v0y = 7.67 m/s

This is where I get stuck at because when I try to go back and plug in v0 = v0y into the third x equation, I'm still left with two variables, θ and v0. I don't know any other way to approach this formula. I searched online and came across a similar problem that used two different formulas I've never seen before:

y = [(v0)sin2θ]/2g

x = [(v0)sin(2θ)]/g

That person said to divide the first formula by the second and solve for angle first, then initial velocity. When I tried that, my answers where θ = 63.4° and v0y = 8.58 m/s.

The book says the correct answers are θ = 58.3° and v0 = 9.76 m/s

Last edited: Sep 5, 2013
2. Sep 5, 2013

### rude man

It looks simple but it's a cut above the average in difficulty.

Start by writing x and y components of velocity and then distance. You should wind up with 2 equations and 3 unknowns.

3. Sep 5, 2013

### TSny

The minimum initial speed that will get the ball over the fence does not necessarily correspond to having the ball be at maximum height as it passes over the fence.

Did you mean to say that the answer is vo = 9.76 m/s rather than voy = 9.76 m/s?

4. Sep 5, 2013

### SammyLP250

Why yes i did :tongue: thank you. And I did not know that, I assumed the ball was at max height at 3m

So I get

v0x = v0cosθ

v0y = vosinθ

and for distance

x = v0xt

y = v0yt + .5(9.81)t2

5. Sep 5, 2013

### rude man

Good so far EXCEPT for the sign in front of the second term. But a suggestion: don't put numbers in until the end. You lose the ability to check for dimensional compatibility in your terms as you go along. These terms get a bit messy as you will see.

So how about x = v0xt
y = v0yt - gt2/2.
I gave you a 'gimmie' by correcting the sign of the 2nd term in y. You do see why, right?

OK, now how about letting y = h and x = d where d = 6m and h = 3m, and letting t = T since this happens at some instant T. Use d and h, not 6 and 3.

6. Sep 5, 2013

### SammyLP250

Gotcha. And I understand why g is negative.

So they will become

d = v0xT

h = v0yT - gT2

7. Sep 5, 2013

### rude man

Right.
Now you're faced with 2 equations and 3 unknowns (vx0, vy0 and T). What does that suggest to you?

8. Sep 5, 2013

### SammyLP250

I'll set T = d/v0x

and substitute T into the second equation.

h = [(v0yd)/v0x] - g (d/v0x)2

So now I'm left with two unknowns!

9. Sep 5, 2013

### rude man

Excellent move! Notice you now have vy0 as a function of vx0.

Next step: what are you trying to minimize?

10. Sep 5, 2013

### TSny

Did you lose a factor of 1/2 in the h equation?

11. Sep 5, 2013

### SammyLP250

I'm stuck right here. But I think I am trying to minimize v0x

12. Sep 5, 2013

### SammyLP250

Ahhh That's right. Thank you again. These small mistakes are going to be the death of me.

h = [(v0yd)/v0x] - [g(d/v0x)2/2]

13. Sep 5, 2013

### TSny

You need to minimize vo. So, write your equation in terms of vo and θo instead of in terms of vox and voy.

14. Sep 5, 2013

### SammyLP250

Alright then so h = [(v0sinθ0d)/(v0cosθ0)] - (g(d/v0cosθ)2)/2

which become h = tanθ0d - g(d/v0cosθ0)2/2

15. Sep 5, 2013

### TSny

Yes, good. If you can solve this for vo as a function of θo, you will have a function for which you want to find the minimum.

16. Sep 9, 2013

### SammyLP250

I'm sorry I haven't replied back. It turns out I wasn't the only one in the class that had trouble with this problem and the teacher realized that and worked it out in class the next day. After looking over it a couple times, I understand what to do now. This problem is definitely not as simple as it looks.

But thank you all for helping me along the way, cause Lord knows I would have just sat there looking at it for hours lol

17. Sep 9, 2013

### TSny

OK. I hope you enjoy your class.