# Projectile motion (soccer kick)

1. Dec 29, 2008

### boomer77

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

The soccer goal is 26.19 m in front of a soccer player. She kicks the ball giving it a speed of 20.77 m/s at an angle of 21.20 degrees from the horizontal. If the goalie is standing exactly in front of the net, find the speed of the ball just as it reaches the goalie.

2. Relevant equations

3. The attempt at a solution

I am unsure how to set up the problem

I know dx= 26.19m i think my dy= 0 because i'm pretty sure this is a symmetrical trip

is the 20.77 m/s the Vx or Viy?

a=-9.8

and right now i do not have t

is there any flaws in what i am sure of and can anyone clarify in the regions i am unsure of?
thanks!

2. Dec 29, 2008

It is at an angle $\theta=21.2$ so it is both. You must resolve v into a horizontal and vertical component.

Once you do this, t can be eliminated using two different equations.

3. Dec 29, 2008

### boomer77

so my vx will be 20.77cos(21.2 and my viy will be 20.77sin(21.20?

explain how you'd eliminate time, i don't see how you'd do that

4. Dec 29, 2008

You have:

1. $v_y^2=(v_y)_o^2+2a_y(y-y_o)$
2. $y-y_o=(v_y)_ot+\frac{1}{2}a_yt^2$
3. $v_y=(v_y)_o+a_yt$
4. $x-x_o=(v_x)t$

Right? So you can either eliminate t or solve for t and use that value elsewhere. Whichever you prefer.

I would suggest that the first thing that you do when solving Projectile Motion Problems is write down the 4 equation above at the top of your paper. Then you can check to see which of the variables you are given and which you need to find.

Casey

Last edited: Dec 29, 2008
5. Dec 29, 2008

### boomer77

okay, but i haven't ever seen the equations like that before. i'm pretty new to this material.

so could i go about the problem like this:

dx= 26.19m dy= 0 (it is a symmetrical trip right?)
vx= 20.77cos21.2 viy=20.77sin21.2
t=26.19/20.77cos21.2 vf= ?
a= -9.8 m/s^2

then use vf=vi+at
vf= 20.77sin21.2+ -9.8(26.19/20.77cos21.2)
vf= -5.74m/s

sqrt vx^2+vfy^2

but on my problem set my answer keeps coming up wrong is there an error somewhere?

6. Dec 29, 2008

Okay, I am having trouble following your work. But here is the idea. (notice I numbered the equations in post #4)

Use equation 4 to find t which you did. Now use t and equation 2 to find $v_y$. Note that vx does not change. Now you should be able to find V using Pythagorean.

What are you using for Vx? I am getting a number slightly larger than your value of V.

7. Dec 29, 2008

### boomer77

could you show me how to use that equation? i know how to do the vyt+1/2 at^2 it's the y-y0 part i don't get.

8. Dec 29, 2008

That's just displacement! It's what you have as dy It's zero.

Get to know those equations, they are the standard. I don;t know what your professor has been using, but he/she should be slapped for not using the conventional eqs.

9. Dec 29, 2008

### boomer77

oh sorry. i'm in highschool so looking at those equations are a little confusing

these are the equations my teacher showed us for projectile motion they're probably what you are showing me just simplified

x direction y direction
d= vt d=vit+1/2at^2
vf=vi+at
d= (vi+vf/2)t

could you tell me what one of these equations to use? i know that they're probably the same as what you showed but this is what i've been taught and it would help me a little better. i apologize for any frustration

10. Dec 29, 2008

### boomer77

sorry for the run on of equations that got mushed together

here's a redo
x direction
-d=vt

y direction
-d=vit+1/2at^2
-vf=vi+at
-d= (vi+vf/2)t

11. Dec 29, 2008

### rl.bhat

dx= 26.19m dy= 0 (it is a symmetrical trip right?)

This is not true.
With angle 21.2 degree and initial velocity 20.77, the range of the projectile is 29.68 m.
( Use the formula R = Vo*cos(theta)*2Vo*sin(theta)/g )

Now Vx = 20.77*cos(21.2) = 19.36m/s and Vyf = -5.74 m/s. Find the velocity.

12. Dec 29, 2008

### boomer77

i get 18.4895...
should i round this number or leave it?

13. Dec 29, 2008

### rl.bhat

i get 18.4895...
should i round this number or leave it?

It is wrong.
V = [ 19.36^2 +(- 5.74)^2]^1/2 = [19.36^2 + 5.74^2]^1/2

14. Dec 29, 2008

### boomer77

i now have 184.5348

am i doing something wrong?

15. Dec 30, 2008

### rl.bhat

Yes. Check the calculation again. The answer cannot be that much large.

[19.36^2 + 5.74^2]^1/2 .......> sqrt.[19.36^2 + 5.74^2]

16. Dec 30, 2008

### boomer77

do i subtract that because now i have 32.95

17. Dec 30, 2008

### rl.bhat

V = [ 19.36^2 +(- 5.74)^2]^1/2 = [19.36^2 + 5.74^2]^1/2 = 20.19 m/s

18. Dec 30, 2008

Which part is not true? The displacement is the x direction is 26.19m and the displacement in y=0. I do not see the need for range formula

19. Dec 30, 2008

### boomer77

oooohhhh i see what i did wrong! thanks a bunch for the help!

20. Dec 30, 2008

### rl.bhat

Which part is not true? The displacement is the x direction is 26.19m and the displacement in y=0. I do not see the need for range formula
You are right. In this problem range formula is not needed. But at x = 26.19 m, y is not equal to zero as boomer77 mentioned. To show that I used the range formula.