How Fast Must a Soccer Player Run to Catch a Ball?

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Homework Statement



A soccer player kicks a ball to his teammate, who is
a distance d away. Even though the kick launches the
ball with speed v0 and angle θ0 , the teammate knows it
will not travel far enough to reach him before it lands.
So as soon as the ball is kicked, the teammate begins
running toward the ball. If he is to meet the ball just
before it hits the ground, show that his average speed
must be vp = (gd / 2v0 sin θ0) - v0 cos θ0
where g is the acceleration due to gravity. Neglect air
resistance.



Homework Equations



parabolic motion equation

The Attempt at a Solution



I understand that t0 = tp and that the distance that the ball travels depends on the angle at which it is kicked. I solved for y to describe the trajectory of the ball in terms of angle theta and I know that xp = vp*t , then my plan was to solve the trajectory equation for x and use the fact that x0 + xp = d but I can't solve the trajectory equation for x since it quadratic. What do I do?
 
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Hi Stochastic13! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)
Stochastic13 said:
… but I can't solve the trajectory equation for x since it quadratic.

What is your trajectory equation for x ? :smile:
 
Thank you, I feel really grateful that we have a forum like this and people such as yourselves to help us first year physics students. The equation that I'm trying to solve for x is:
y = (tan[tex]\theta[/tex]) x - g/(2(v0) (sin[tex]\theta[/tex])2) x2 which describes the parabolic trajectory of a particle--in this case a soccer ball.
 
Hi Stochastic13! :smile:

(have a theta: :wink:)

θ is known, isn't it, and y = 0?

ok, then your equation is 0 = Ax - Bx2, so x = 0 or A/B. :wink:
 
I'm sorry, I don't understand why y should = 0. Isn't the ball kicked in the air at angle theta? And if the equation becomes 0 = A x - B x2 doesn't x = -2A/B and not A/B? Also in the problem I have to show that the velocity of player has to equal (gd / 2v0 sin θ0) - v0 cos θ0 how can I do that if I reduce the quadratic equation to constants A and B?

P. S. Thanks for the symbols to paste.
 
Stochastic13 said:
I'm sorry, I don't understand why y should = 0. Isn't the ball kicked in the air at angle theta?

I haven't seen all your equations, but I'm assuming that y is height, and that you're solving for when the ball returns to the ground, ie y = 0.
And if the equation becomes 0 = A x - B x2 doesn't x = -2A/B and not A/B?

Nooo :redface: …

0 = A x - B x2 = x(A - Bx). :wink:
 
Oh, I see... I don't know I tried to use the value I get for x0 , when I solved the quadratic, and plug that into the equation x0 + xp = d which is (cosθ sinθ (2v0)/g) - x0/v0cosθ = d and i get -1 = (gd / 2v0 sin θ0) - v0 cos θ0 which is close to what I'm trying to show but still not quite there.
 
Tiny-tim thanks a lot for the hint, it turned out that it's exactly what I needed to solve this problem. I made a mistake earlier in my algebra, but I took a look at my derivation again and it worked out beautifully.