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.
Posts: 3060 | Registered: Nov 2003
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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.
Posts: 2926 | Registered: Sep 2005
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posted
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.
Posts: 10645 | Registered: Jul 2004
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posted
^^ --> I'm just so used to solving the quadratic formula that I did it again without thinking about the simpler way ^^.
Posts: 3060 | Registered: Nov 2003
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quote:Originally posted by Juxtapose: is this a frictionless problem?
I imagine that it would be hard to kick a football in a frictionless situation. You'd probably land flat on your butt.
Posts: 10397 | Registered: Jun 2005
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