# Homework Help: Projectile motion problem

1. Jan 19, 2005

### nineeyes

Problem : find a $$v_i$$ and angle $$\theta$$ that will be able to hit the target 795 m away and clear the 60 m hill 600 m away. I attached a picture of the problem.

I was trying to do this using the equation
$$y=y_0+(x-x_0)*tan(\theta)-\frac {1}{2}g (\frac {x-x_0}{v_0*cos(\theta)})^2$$

I used this by seperating the problem into two parts from the origin to x=600 m and y=60m ($$y_0 = x_0 = 0$$) and the second part from x= 795 m and y=0 m($$y_0=60m , x_0=600m$$)
I know this is wrong, because I couldn't get an answer. I was wondering what the best way to approach this problem would be. Thanks in advance!

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Last edited: Jan 20, 2005
2. Jan 20, 2005

### Andrew Mason

Try using your equation and finding the expressions for $v_0$ and $\theta$ where (x,y) = (600,60) and (x,y) = (795,0)

AM

3. Jan 20, 2005

### nineeyes

Hi,
I tried to do this before, I couldn't solve for either variable and when I plug it into my calculator to solve it says that it's false. This is where I got stuck. Is there another way to solve for it or am I doing something wrong?
$$60=600*tan(\theta) - \frac{1}{2}*g*\frac{600}{v_0*cos(\theta)}$$
and
$$0=795*tan(\theta)-\frac{1}{2}*g*\frac{795}{v_0*cos(\theta)}$$

Last edited: Jan 20, 2005
4. Jan 20, 2005

### Andrew Mason

In the last term, t (= x/vcos) should be squared:

$$60=600*tan(\theta) - \frac{1}{2}*g*\frac{600^2}{v_0^2cos^2(\theta)}$$
and
$$0=795*tan(\theta)-\frac{1}{2}*g*\frac{795^2}{(v_0^2cos^2(\theta))}$$

Substituting $t=x/v_0cos(\theta)$ into the general equation of motion, you get the general solution to the quadratic for the range, R:

$$R = v_0^2sin^2\theta/2g$$

Then substitute that into the equation of motion to get:
$$y = xtan\theta - \frac{1}{4R}tan^2\theta$$ where R = range (795)

By substituting x,y = 600,60 you have a quadratic equation in terms of tan(theta) only. You will have to check my algebra - I am not guaranteeing that I haven't missed something.

AM

Last edited: Jan 20, 2005
5. Jan 20, 2005

### nineeyes

Hi,
Someone asked my professor about this in class and he said we only need to use the equation $$y=y_0+(x-x_0)*tan(\theta)-\frac {1}{2}g (\frac {x-x_0}{v_0*cos(\theta)})^2$$ Once.
He said that we need to put $$v_0$$ in terms of $$\theta$$. and find a $$\theta$$ that would give a minimum value of $$v_0$$.
$$v_0=sqareroot((\frac{.5*g*x^2}{x*tan(\theta)(cos^2(\theta))}))$$
using the (795,0) and $$x_0 = y_0 = 0$$
I was trying this but I dont recall how to find an angle that would give a minimum value. Can you offer any help regarding this ?
Thanks

Last edited: Jan 20, 2005
6. Jan 20, 2005