This is topic Physics? Me on knees, peeza, please? in forum Books, Films, Food and Culture at Hatrack River Forum.


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Posted by Phanto (Member # 5897) on :
 
*sigh*

I was reviewing my notes from way back when, and realized that I had not done a few problems...I decided to do them, feeling it should be easy, considering my now more extensive knowledge of physics.

So the problem is:

A football is kicked at 55 degrees above the horizontal, 3 feet from the ground. It travels 618 ft. What was the velocity with which it was kicked?

Easy! Just write down the unknowns...

VCosα = Vx
VSinα = Vy
T = time
d = distance, 618 ft
α = 55 degrees
h = height when hits ground; -3 ft from starting position

AND

VxT = d

y(T) = VyT - .5gT^2
gT^2-2VyT+2h = 0

So I use the quadratic formula, solve for T, then plug into the VxT = d equation, but then can't really solve that equation...

So where am I going wrong? Is there some simplier method of doing this? Please, any hints/tips welcome. I already have the answer, if that is of any use.
 

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Posted by SenojRetep (Member # 8614) on :
 
Can you solve it like this:

x = Vx*t so 618 = V*cos(55)*T (where T is the time at which it touches down). Therefore V*T = 618/cos(55).

Now substitute that into your y equation: 0 = 3+V*sin(55)*T-.5gT^2 or, after substitution, 0 = 3+618*tan(55)-.5gT^2. Therefore T = sqrt(6/g+618/g*tan(55)) and V = 618/cos(55)/sqrt(6/g+618/g*tan(55)). I get 205 ft/s.
 

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Posted by Juxtapose (Member # 8837) on :
 
618 feet? yowza, that's some kick.

EDIT - is this a frictionless problem?
 

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Posted by King of Men (Member # 6684) on :
 
Your method will give the correct result, and I can't offhand think of a simpler one (there again, my brain is mush at the moment). The equation for VxT = d looks hairy, but it turns out fairly nice if you just persevere. Move the non-square-root terms over, square both sides to get rid of the nasty square root, and simplify from there.
 

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Posted by Tatiana (Member # 6776) on :
 
It travels 618 ft.

Does that mean the horizontal distance, as I assume? Not the total path length?

Knowns:
theta = 55 degrees
y0 = 3 feet
x0 = 0
x(Tfinal) = 618 feet
y(Tfinal) = 0

A: y(t) = y0 + v0y * t - 1/2 a t^2
B: vy(t) = v0y - a * t
a = 32 ft/sec^2
C: x(t) = x0 + v0x * t

what we are trying to find:
v0

More things we know:
D: v0x = v0 * cos 55 degrees
E: v0y = v0 * sin 55 degrees

Oh, and we also know that at the peak of it's arc, vy(Tpeak) = 0

[ May 01, 2006, 10:02 PM: Message edited by: Tatiana ]
 

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Posted by Tatiana (Member # 6776) on :
 
So lets see what we get when we put in zero for vy(t) in equation B:.

vy(t) = 0 = v0y - a * t = v0y - 32 * Tpeak
from this we get:

F: v0y = 32 * Tpeak
 

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Posted by Phanto (Member # 5897) on :
 
Yeah, no friction.

I tried it again, and instead of using the quadratic formula, solved for T in terms of Vx, then plugged it into the quadratic.

That solved the problem .

The answer is about 145 ft/s.
 

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Posted by Tatiana (Member # 6776) on :
 
Awesome! I won't finish, then.
 

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Posted by Phanto (Member # 5897) on :
 
^^ --> I'm just so used to solving the quadratic formula that I did it again without thinking about the simpler way ^^.
 

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Posted by Tante Shvester (Member # 8202) on :
 

quote:

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Originally posted by Juxtapose:
is this a frictionless problem?

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quote:

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Originally posted by Tante Shvester:

quote:

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Originally posted by Juxtapose:
is this a frictionless problem?

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I imagine that it would be hard to kick a football in a frictionless situation. You'd probably land flat on your butt.

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