- #1
bornofflame
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Homework Statement
A certain football placekicker can kick the ball a distance d if she kicks it at 45°.
a. Find an equation for the velocity of the kick. (You may use the range equation R = (V02/g)sin2θ.)
b. Now imagine she kicks at a different angle θ. Show that a formula for the height y at each horizontal position x from where she kicks is y = xtanθ - (1/2d)x2(secθ)2.
(Hint: This is a projectile motion problem, find x(t), and y(t) just as we did in finding the range equation and put them together to get y(x).
c. Find the angle that gives the maximum height at a given distance x. (Hint: In calculus how did you maximize a value?)
Homework Equations
R = (V02/g)sin2θ
y = xtanθ - (1/2d)x2(secθ)2
The Attempt at a Solution
a. V = ?
V02 = Rg/sin2θ
V0 = √(Rg/sin2θ)
V = V0 + at = √(Rg/sin2θ) + at
Final answer: V = √(Rg/sin2θ) + at
b. x(t), y(t) = ?
x = x0 + v0t + ½at2
Using a triangle where θ is the angle between d and x
we can see the following relationships:
x = dcosθ
y = dsinθ
d = x / cosθ
y = (x / cosθ) sinθ = xtanθ This looks like the first half of the equation that we are trying to get to, but I'm not quite sure how to get the other part.
If I look at this in terms of a Pythagorean triangle:
d2 = x2 + y2
y2 = d2 - x2 = (x/cosθ)2 - x2 = x2/sec2θ - x2 = x2(sec2θ - 1) = x2tan2θ = xtanθ
Which kinda makes me feel better as it results in the same thing and gives me the impression that I'm on the right track at least.c. y = xtanθ - (1/2d)x2(secθ)2
Set the derivative to 0 and solve for θ
y = xtanθ - (1/2d)x2(secθ)2 = 0
y' = xsec2θ - (1/2d)x22sec2θtanθ
= xsec2θ - (1/d)x2sec2θtanθ
At first I was thinking that θ would have to be π/2 due to sec2θ being present in both parts of the equation, but the only way for that to work is by taking the limit as θ → π/2 which isn't, I think, the same. So, simplifying further, I got this:
xsec2θ(1 - (x/2d)θtanθ)
Where I still have the d and x haunting me. That is where I'm not sure which way to go. I'm thinking that I may need to solve part b, before I can actually tackle this section.