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## Homework Statement

A picture of the problem can be seen http://images3a.snapfish.com/232323232fp733;3>nu=52::>379>256>WSNRCG=335;<6;737347nu0mrj".

Original question:

An archer is standing inside a long building whose ceiling is 11 m high. The archer can shoot and arrow at a maximum velocity of 62 m/s, What is the angle that he should shoot at to acheive the greatest range inside the building? How long does the building have to be for the arrow to land inside? How long is it in the air? How fast is it going just before it hits the ground?

## Homework Equations

[tex] x_{f}=\frac {a t^{2}}{2} + V_{0}t+x_{o} [/tex]

and other motion equations.

## The Attempt at a Solution

Known/Given:

[tex] V_{0}=62 \frac {m}{s} [/tex]

[tex] y_{max}=11m [/tex]

[tex] y_{i}=x_{i}=0 [/tex]

So motion in the x direction is simply going to be:

[tex] x_{f}=V_{0}cos(\theta)t [/tex]

and the y direction should be:

[tex] y_{f} = -\frac {gt^{2}}{2}+V_{o}sin(\theta)t [/tex]

where g is the acceleration due to gravity. Now we know that there is going to be a maximum for y at 11 but we need to get this in terms of 1 variable not two. If I take the derivative of the y equation I get:

[tex] \frac {dy_{f}}{dt} = -gt+V_{0}sin(\theta) [/tex]

we can then solve this to get t in terms of theta, also I believe this is actually a partial derivative:

[tex] \frac {\partial y_{f}}{\partial t} = -gt+V_{0}sin(\theta) [/tex]

[tex] t=\frac {V_{0}sin(\theta)}{g} [/tex]

Now we can substitute this back into the position function to get the position function in terms of only theta:

[tex] y_{f}=v_{0}sin(\theta)(\frac{V_{0}sin(\theta)}{g})- \frac {g}{2}(\frac{V_{0}sin(\theta)}{g})^{2} [/tex]

[tex] y_{f} = \frac {V_{0}^{2}sin^{2}(\theta)}{g} -\frac {g}{2}(\frac {V_{0}^{2}sin^{2}(\theta)}{g^{2}}) [/tex]

[tex] y_{f} = \frac {V_{0}^{2}sin^{2}(\theta)}{g} -\frac {V_{0}^{2}sin^{2}(\theta)}{2g} [/tex]

combine the fractions to get:

[tex] y_{f}= \frac {V_{0}^{2}sin^{2}(\theta)}{2g} [/tex]

now set y

_{f}to 11 and solve for theta to get the maximum angle:

[tex] \theta = sin^{-1}(\sqrt{\frac{22g}{V_{0}^{2}}}) \approx 13.6993 [/tex]

Now to figure out how long the building needs to be we should simply need to use the t=F(θ) equation for the new angle and then use that t in the x equation of motion.

[tex] t=\frac {V_{0}sin(\theta)}{g} = \frac {62sin(13.6993)}{9.8} \approx 1.498 seconds[/tex]

[tex] x_{f}=V_{0}cos(\theta)t=62cos(13.6993)(1.498) \approx 90.23 meters [/tex]

I think this is wrong though as the time there is only the time to y max not the total flight time, If you use the time found below for total flight time of the arrow you will get:

[tex] x_{f}=V_{0}cos(\theta)t=62cos(13.6993)(2.996) \approx 180.5 meters [/tex]

To find how long it is in the air we should take the y equation of motion and solve it for t knowing theta and yf (which is 0):

[tex] y_{f}=\frac {1}{2}gt^{2}+V_{0}sin(\theta)t [/tex]

I will use complete the square here because i find it the easiest method for solving quadratic equations:

[tex] 0 =- \frac {1}{2}g(t-\frac{62sin(\theta)}{g})^{2}+ \frac {62^{2}sin^{2}(\theta)}{19.6} [/tex]

solving this for t we get 2.996 seconds.

Finally for how fast is it going when it hits the ground, I think that they want the magnitude of the velocity vector which is:

[tex] \bar{V} = \sqrt{V_{x}^{2}+V_{y}^{2}} [/tex]

and

[tex] V_{x} = V_{0}cos(\theta) \approx 60.236[/tex]

[tex] V_{y} = -gt+V_{0}sin(\theta) \approx -14.68[/tex]

then

[tex] \bar{V}= \sqrt{ 60.236^{2}+(-14.68)^{2}} \approx 62 \frac {m}{s} [/tex]

Is this the correct way to solve the problem?

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