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Lucky12213's question at Yahoo! Answers regarding projectile motion

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MarkFL

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Feb 24, 2012
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Here is the question:

I need help with Trig/Physics?

My Trigonometry class is currently working on some physics related math and I have a hard time understanding how to do the problem. My teacher said the textbook did not explain it properly and to use the notes we took instead. I wasn't there for the notes so again my understanding decreases. The problem is as follows:

"Andrew is attempting the longest field goal of his career. The field goal post is a distance of 20 meters from where he will kick the ball, and the cross bar on the post is 2 meters above the ground. Will Andrew make the field goal if he kicks the football with enough force to give it an initial velocity of 15 m/s at an angle of 45 degrees above the ground? By what height does he make or miss the field goal?"

If someone could help me out and guide me through the problem solving process I would really appreciate it. Thank you!
Here is a link to the question:

I need help with Trig/Physics? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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MarkFL

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Feb 24, 2012
13,775
Hello Lucky12213,

I would begin with the parametric equations of motion:

(1) \(\displaystyle x=v_0\cos(\theta)t\)

(2) \(\displaystyle y=-\frac{g}{2}t^2+v_0\sin(\theta)t\)

To eliminate the parameter $t$, we solve (1) for $t$ and substitute into (2) to get:

\(\displaystyle y=\tan(\theta)x-\frac{g}{2v_0^2\cos^2(\theta)}x^2\)

Now we have the height $y$ of the ball as a function of the initial velocity $v_0$, the launch angle $\theta$ and the horizontal displacement $x$. Distances are in meters.

Using the given values:

\(\displaystyle v_0=15,\,\theta=45^{\circ},\,x=20,\,g=9.8\)

we find:

\(\displaystyle y=1\cdot20-\frac{9.8}{2\cdot15^2\cdot\left(\frac{1}{\sqrt{2}} \right)^2}20^2=20-\frac{9.8\cdot20^2}{15^2}=\frac{116}{45}\approx2.58\)

So, we see the ball clears the goal post by \(\displaystyle \frac{26}{45}\,\text{m}\).

To Lucky12213 and any other guests viewing this topic, I invite and encourage you to post other projectile problems in our Other Topics forum.

Best Regards,

Mark.