Projectile motion problem (find the angle) other ways?

[powerof]

Homework Statement

A soccer player is taking a penalty kick from a distance of 11 meters from a goal which is 2.44 m high (forgive the possible incorrect use of English, since I'm not a native speaker). Given an initial speed of the ball of 90 km/h (which is 25 m/s²) calculate the initial upward angle of the ball for it to hit the upper bar (i.e., have height = 2.44 after having traveled 11 m horizontally).

Homework Equations

The trajectory equation for [itex]x_0=y_0=0[/itex] (initial horizontal and vertical positions) is:

[itex]y = (tan\theta)x - (\frac{g}{2(v_0)^2 cos^2 \theta})x^2[/itex]

Some known constants in the above equation are:

[itex]v_0=25 m/s \ \ \ \ \ \ g = 9.81 m/s^2[/itex]

And obviously [itex]\theta\in(0,\frac{\pi}{2})[/itex]

The Attempt at a Solution

The straightforward way is to just substitute the known data in the main equation and solve for [itex]\theta[/itex]:

[itex]2.44 = 11(tan\theta) - (\frac{9.81}{2(25)^2 cos^2\theta})11^2[/itex]

But I'm lazy and I find solving for theta to be long and tedious (for me at least) so what I'm asking is:

Are there are any other ways for finding the angle in the mentioned problem besides the classic method I gave you?

I feel it's a long shot but I'm in no hurry and it doesn't hurt to ask. Also it doesn't necessarily have to be an easier method. Just post any you can think of.

Thanks for reading this and hopefully answering it.

Have a nice day!

 

[SammyS]

Homework Statement

A soccer player is taking a penalty kick from a distance of 11 meters from a goal which is 2.44 m high (forgive the possible incorrect use of English, since I'm not a native speaker). Given an initial speed of the ball of 90 km/h (which is 25 m/s²) calculate the initial upward angle of the ball for it to hit the upper bar (i.e., have height = 2.44 after having traveled 11 m horizontally).

Homework Equations

The trajectory equation for [itex]x_0=y_0=0[/itex] (initial horizontal and vertical positions) is:

[itex]y = (tan\theta)x - (\frac{g}{2(v_0)^2 cos^2 \theta})x^2[/itex]

Some known constants in the above equation are:

[itex]v_0=25 m/s \ \ \ \ \ \ g = 9.81 m/s^2[/itex]

And obviously [itex]\theta\in(0,\frac{\pi}{2})[/itex]

The Attempt at a Solution

The straightforward way is to just substitute the known data in the main equation and solve for [itex]\theta[/itex]:

[itex]2.44 = 11(tan\theta) - (\frac{9.81}{2(25)^2 cos^2\theta})11^2[/itex]

But I'm lazy and I find solving for theta to be long and tedious (for me at least) so what I'm asking is:

Are there are any other ways for finding the angle in the mentioned problem besides the classic method I gave you?

I feel it's a long shot but I'm in no hurry and it doesn't hurt to ask. Also it doesn't necessarily have to be an easier method. Just post any you can think of.

Thanks for reading this and hopefully answering it.

Have a nice day!

Are you too lazy to solve a quadratic equation?

To get to that point will require using a couple of trig identities and doing some basic algebra.

 

[barryj]

Well, if you have a graphing calculator like a TI84, you might be able to plot the function and look for a zero.

 

[powerof]

Are you too lazy to solve a quadratic equation?

To get to that point will require using a couple of trig identities and doing some basic algebra.

I guess someone should've told me that before playing with sines and cosines for half an hour. Anyway, I looked it up on the Internet (and this post is part of my 'research') and I suppose you're referring to this.

I'll have to admit that it would not have occurred to me to substitute 1/cos^2 with tan^2 + 1 but once you do that the rest is just algebra, as you say.

Thank you very much for the hint.

 

[voko]

This could have been solved almost without any trig by considering the vertical and horizontal motions separately.

 

[powerof]

This could have been solved almost without any trig by considering the vertical and horizontal motions separately.

This made me curious. How so?

Given: [itex]v_y = v_0 sin \theta \ \ \ \ \ v_x = v_0 cos \theta[/itex]

Vertical motion: [itex]y = y_0 + v_y t + \frac{1}{2}at^2 \ \ \ \ \rightarrow y = v_0 (sin \theta)t - \frac{1}{2}g t^2[/itex]

Horizontal motion: [itex]x = x_0 + v_x t + \frac{1}{2}at^2 \ \ \rightarrow x = v_0 (cos \theta)t[/itex]

How would you have solved it differently?

 

[voko]

Stick with ##v_x## and ##v_y## till the end. When you find them, you will obtain ##\theta##, this being the only time you have to use trigonometry, but it will be of a most basic kind.

 

[Basic_Physics]

Shoot the monkey

Maybe this will also do?

 

[barryj]

I think powerof's equations are correct.

 

[verty]

I think this problem may be just inherently difficult.

To get a general formula for ##\tan{θ}##, make the substitution ##k = \frac{v_0^2}{g}##. After much algebra, a not too hideous formula results.