Physics problem, hairy algebra

**TwoPlusTwo** · February 4th 2011, 02:58 AM

We have a mini catapult that fires balls at a velocity of 5 m/s. It is placed 1 meter above the ground and pointed at an angle $\theta$ towards the positive x-axis. What angle(s) do we need to set it to in order to hit a target on the ground at x=2.7 meters?

In other words:

$v_{0}=5 m/s$
$x(t)=2.7 m$
$y(t)=-1 m$ (assuming the catapult is standing in the origin)
$\theta=?$

My thinking so far

First express the velocity in the x-direction and y-direction separately:

$v_{x}=v_{0}cos\theta$

$v_{y}=v_{0}sin\theta-gt$

Then find a way to express t in terms of the angle and the positions:

$x(t)=\frac{v_{x}}{t}$

$t=\frac{v_{x}}{x(t)}=\frac{v_{0}cos\theta}{x(t)}$

Use what I now have to write y as a function of theta:

$y(t)=v_{y}-\frac{1}{2}gt^2$

$y(t)=v_{0}sin\theta-gt-\frac{1}{2}gt^2$

$y(\theta)=v_{0}sin\theta-g(\frac{v_{0}cos\theta}{x(t)})-\frac{1}{2}g(\frac{v_{0}cos\theta}{x(t)})^2$

Since I know x, y and v, I should now be able to solve for theta.

Have I set this up correctly? Is there maybe an easier way to calculate the angle? And finally, if this is the best way to find it, how the hell do I solve for theta??

**Unknown008** · February 4th 2011, 04:21 AM

Um... the equation for the $v_y$ gives the actual vertical velocity at time t, not the vertical component of the initial velocity.

And then, you'd also like to get the displacement of the ball in terms of vertical and horizontal components.

Along the horizontal, you have:

$s_x = v_o\cos\theta t$

Vertically;

$s_y = s_o + v_o \sin\theta t - \dfrac12 gt^2$

Horizontally, $s_x = 2.7 m$

Vertically, $s_y = -1 m$

You should be able to get the angle more easily like this.

**TwoPlusTwo** · February 4th 2011, 05:13 AM

So I'm supposed to get one angle from each equation by substituting in my expression for t?

$t=\frac{v_{x}}{s_{x}}=\frac{v_{o}cos\theta}{s_{x}}$

I did that, and got the 57 degrees from the first one, but the second one I still can't solve..

**Unknown008** · February 4th 2011, 05:19 AM

Um...

The two equations that you get are:

$-1 = 0 + 5 \sin\theta t - \dfrac12 gt^2$

$2.7 = 5 \cos\theta t$

From that you get:

$t = \dfrac{2.7}{v_x} = \dfrac{2.7}{5\cos\theta}$

Hm... this will be a little more difficult to solve than I expected, since you get a quadratic. I forgot the -1 would be in the quadratic.

EDIT:

Also remember that Speed = Distance/Time.

So, this : t = v/s is false, but t = s/v is true.

**BAdhi** · February 4th 2011, 05:31 AM

$-1 = 0 + 5 \sin\theta t - \dfrac12 gt^2$ .................(1)

$t = \dfrac{2.7}{v_x} = \dfrac{2.7}{5\cos\theta}$ ..................(2)

substitute t on the (1) from (2)

you'll find a quadric equation of the variable $\tan\theta$

**TwoPlusTwo** · February 4th 2011, 05:41 AM

Great! Thanks guys. Finally I know what equation I need to solve. I'll give it a shot right away.

**skeeter** · February 4th 2011, 06:20 AM

$\Delta x = v_0 \cos{\theta} \cdot t$

$t = \dfrac{\Delta x}{v_0 \cos{\theta}}$

$\Delta y = v_0 \sin{\theta} \cdot t - \frac{1}{2} gt^2$

$\Delta y = v_0 \sin{\theta} \cdot \dfrac{\Delta x}{v_0 \cos{\theta}} - \frac{1}{2} g \left(\dfrac{\Delta x}{v_0 \cos{\theta}}\right)^2$

$\Delta y = \Delta x \tan{\theta} - \dfrac{g \Delta x^2}{2v_0^2} \sec^2{\theta}$

$\Delta y = \Delta x \tan{\theta} - \dfrac{g \Delta x^2}{2v_0^2} (1 + \tan^2{\theta})$

$0 = -\left(\Delta y + \dfrac{g \Delta x^2}{2v_0^2}\right) + \Delta x \tan{\theta} - \dfrac{g \Delta x^2}{2v_0^2}\tan^2{\theta}$

$a = - \dfrac{g \Delta x^2}{2v_0^2}$

$b = \Delta x$

$c = -\left(\Delta y + \dfrac{g \Delta x^2}{2v_0^2}\right)$

use the quadratic formula to solve for $\tan{\theta}$ ... you will get two valid solutions.

inverse tangent yields $\theta \approx 60^\circ$ and $\theta \approx 10^\circ$

**Unknown008** · February 4th 2011, 06:20 AM

Anyway, the problem remains tough to solve...

After simplifying we get:

$-1.42884 + 2.7\sin\theta\cos\theta + \cos^2\theta = 0$

Spoiler:

EDIT: How can I forget 1/cos^2A = sec^2A = 1 + tan^2A

**TwoPlusTwo** · February 4th 2011, 06:54 AM

I can see you already posted the answer Unknown (edit: and Skeeter), but hey I needed to practice my LaTex skills anyways, so here it goes:

$- \dfrac12 g(\dfrac{2.7}{5\cos\theta})^2+5 \sin\theta (\dfrac{2.7}{5\cos\theta})+1=0$

$<br /> (-\frac{1}{2}(2.7)^2(\frac{1}{5^2})\frac{1}{cos^2\th eta}+2.7tan\theta+1=0$

$-1.43(sec^2\theta)+2.7tan\theta+1=0$

$-1.43(1+tan^2\theta)+2.7tan\theta+1=0$

$-1.43tan^2\theta+2.7tan\theta-0.43=0$

Using the quadratic formula I get:

$tan\theta=1.71$ and $tan\theta=0.176$

Which gives

$\theta=59.7^\circ$

and

$\theta=10^\circ$

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