MHB Lucky12213's question at Yahoo Answers regarding projectile motion

AI Thread Summary
The discussion centers on a physics problem involving projectile motion, where Andrew attempts a field goal from 20 meters away with an initial velocity of 15 m/s at a 45-degree angle. The solution involves using parametric equations to determine the height of the ball at the goal post. By substituting the given values into the equations, it is calculated that the ball reaches approximately 2.58 meters, clearing the 2-meter goal post by about 0.58 meters. The response encourages further questions on projectile motion in a dedicated forum. This highlights the importance of understanding the mathematical principles behind projectile motion in physics.
MarkFL
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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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Hello Lucky12213,

I would begin with the parametric equations of motion:

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

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

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

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

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:

$$v_0=15,\,\theta=45^{\circ},\,x=20,\,g=9.8$$

we find:

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

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

To Lucky12213 and any other guests viewing this topic, I invite and encourage you to post other projectile problems in our http://www.mathhelpboards.com/f22/ forum.

Best Regards,

Mark.
 
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