Lucky12213's question at Yahoo Answers regarding projectile motion

In summary, the person is asking for help with a physics problem involving a football being kicked at a certain velocity and angle to make a field goal. The responder provides a summary of the problem solving process and invites others to post more problems for help.
  • #1
MarkFL
Gold Member
MHB
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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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  • #2
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 http://www.mathhelpboards.com/f22/ forum.

Best Regards,

Mark.
 

Related to Lucky12213's question at Yahoo Answers regarding projectile motion

1. What is projectile motion?

Projectile motion is the motion of an object through the air or space that is affected only by the force of gravity. It follows a parabolic path, meaning the object moves both horizontally and vertically.

2. How is projectile motion calculated?

Projectile motion can be calculated using the equations of motion, which take into account the initial velocity, acceleration due to gravity, and time. It can also be calculated using vector components.

3. What factors affect projectile motion?

The factors that affect projectile motion include the initial velocity, angle of launch, air resistance, and the acceleration due to gravity. These factors can change the trajectory and distance of the object.

4. What is the difference between horizontal and vertical motion in projectile motion?

Horizontal motion is the motion of the object in the x-direction, while vertical motion is the motion in the y-direction. In projectile motion, the horizontal motion is constant, while the vertical motion is affected by the acceleration due to gravity.

5. How is projectile motion used in real life?

Projectile motion is used in many real-life situations, such as throwing a ball, launching a rocket, or shooting a projectile. It is also used in sports like baseball, basketball, and golf, and in engineering to calculate the trajectory of objects.

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