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Projectile Motion football kick Question

  • Thread starter balm
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  • #1
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Can someone help me here?

A kicker can kick a football with an initial velocity of 25 m/s. to score the field goal he must clear the goalpost that is 50 meters away. the crossbarr of the goalpost is 3.44 meters above ground. he kicks the ball from ground level. What two angles can he kick the ball to score?
 

Answers and Replies

  • #2
HallsofIvy
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I can't imagine someone wanting to do a problem like that unless it was homework! I'm moving this to the homework forum.

In the meantime, set up the equations of motion:

x(t)= vxt, y(t)= -g/2 t2+ v_y t
where x(t) and y(t) are the horizontal and vertical distances of the ball from the starting point at time t seconds. vx and vy are the x and y components of the initial velocity (the 25 m/s you are given is initial speed, not velocity). Calculate that from the initial speed, 25 m/s, and the angle which you will need to leave a an unknown. You need to find that angle so that, for some time t, x(t)= 50 and y(t)= 3.44 m.
 
  • #3
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balm said:
Can someone help me here?

A kicker can kick a football with an initial velocity of 25 m/s. to score the field goal he must clear the goalpost that is 50 meters away. the crossbarr of the goalpost is 3.44 meters above ground. he kicks the ball from ground level. What two angles can he kick the ball to score?
I would start by setting up a table of values:

t0=0 s t=?
x0=0 m x=50 m
y0=0 m y=3.44 m
v0x=25cos(theta)
v0y=25sin(theta)
ax=0 m/s2
ay=-g
(I am assuming a coordinate system with origin where the ball was kicked and positive x in the horizontal direction the ball was kicked and positive y upward.)

You've got 4 possible equations to work with, since ax=0:
[tex] x=x_0+v_{0x}t [/tex]
[tex] y=y_0+v_{0y}t+(1/2)a_yt^2 [/tex]
[tex] v_y=v_{0y}+a_yt [/tex]
[tex] v_y^2=v_{0y}^2+2a_y(y-y_0) [/tex]

The goal is to use these equations to write an equation in one unknown. I'll give you a tip: You can't do that here. So try two equations in two unknowns. Again, a tip: Use the first two equations. This will give you two equations in t and theta. It's a complicated problem, but doable. Give it a try from here and see what you can do with it.

-Dan
 
  • #4
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I've just finished doing some work on projectile motion myself :smile: Do you know how to derive or can quote the equation of a path that leaves you with only one variable: tan theta? (you already know x, y, v, and g)...

(I'm working on the principle that the two angles you need to find represent the absolute maximum and minimum values necessary to score...any value between those score anyway)
 
Last edited:
  • #5
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i am trying this but am ending up with one heck of a mathematics mess.
 
  • #6
335
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balm said:
i am trying this but am ending up with one heck of a mathematics mess.
Hint(s): Solve your x equation for cos(theta). Solve your y equation for sin(theta). Now, what is sin^2(theta) + cos^2(theta)? You will eventually get an equation that is called a "biquadratic." i.e. of the form at^4+bt^2+c=0. Let X = t^2, then you have aX^2+bX+c=0 which you can solve for X. Then take the square root to get t. You should now have two (physical) possibilities for t. Use these t values to find theta.

-Dan
 
  • #7
335
4
topsquark said:
Hint(s): Solve your x equation for cos(theta). Solve your y equation for sin(theta). Now, what is sin^2(theta) + cos^2(theta)? You will eventually get an equation that is called a "biquadratic." i.e. of the form at^4+bt^2+c=0. Let X = t^2, then you have aX^2+bX+c=0 which you can solve for X. Then take the square root to get t. You should now have two (physical) possibilities for t. Use these t values to find theta.

-Dan
BTW: One quick check on your answer is that the two theta values add to 90 degrees.

-Dan
 
  • #8
3
0
thanks to all , I solved it. I solved the Vxo equation for time and the put that in the Vyo equation, solved for theta using the quadratic equation and came up with 31 and 62 degrees. Thanks again, I'm sure I will be back again.
 

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