Projectile motion (soccer kick)

[boomer77]

Homework Statement

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.

Homework Equations

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!

 

[Saladsamurai]

Homework Statement

...is the 20.77 m/s the Vx or Viy?

It is at an angle [itex]\theta=21.2[/itex] 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.

 

[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

 

[Saladsamurai]

You have:

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

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

 

[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
to get about 18.5 m/s

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

 

[Saladsamurai]

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 [itex]v_y[/itex]. Note that v_x 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.

 

[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.

 

[Saladsamurai]

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.

 

[boomer77]

oh sorry. I'm in high school 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
vf^2= vi^2+2ad
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

 

[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
-vf^2= vi^2+2ad
-d= (vi+vf/2)t

 

[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.

 

[boomer77]

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

 

[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

 

[boomer77]

i now have 184.5348

am i doing something wrong?

 

[rl.bhat]

i now have 184.5348

am i doing something wrong?

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]

 

[boomer77]

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

do i subtract that because now i have 32.95

 

[rl.bhat]

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

 

[Saladsamurai]

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.

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

 

[boomer77]

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

 

[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.

 

[Saladsamurai]

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.

If I am not mistaken, RANGE=Horizontal displacement=26.19m

...the range of the projectile is 29.68 m

How are you getting this number?

 

[PhanthomJay]

rl.bhat is correct; the range of a projectile kicked from ground level is the horizontal distance traveled by it when it hits the ground., that is, when y=0. In this problem, the ball does not hit the ground at the goalie's feet, rather, it hits her partway up her body (solve for y when it hits the goalie; y is not 0).

 

[Saladsamurai]

rl.bhat is correct; the range of a projectile kicked from ground level is the horizontal distance traveled by it when it hits the ground., that is, when y=0. In this problem, the ball does not hit the ground at the goalie's feet, rather, it hits her partway up her body (solve for y when it hits the goalie; y is not 0).

Interesting point. I have never used the Range formula... how would you know that y not equal to 0 without using range formula? It should be simple enough, but I am not seeing it

 

[PhanthomJay]

QUOTE=Saladsamurai;2017204]Interesting point. I have never used the Range formula... how would you know that y not equal to 0 without using range formula? It should be simple enough, but I am not seeing it [/QUOTE]

Looking in the x direction, and knowing the horizontal distance and horizontal speed, you can quickly solve for t (t= d_x/v_x); and then knowing t, and knowing v_yi =V_i(sintheta) , now solve for y at time t, using y = v_yi(t) -1/2(g) t^2. I think y comes out to a meter + or so, hitting the goalie right around chest level. But the problem never asked for y, it just was looking for v_f, as calculated above.

 

[rl.bhat]

Interesting point. I have never used the Range formula... how would you know that y not equal to 0 without using range formula? It should be simple enough, but I am not seeing it

If you get the final velocity equal to the initial velocity, then y=o.