# Two Projectile Motion Problems

1. Jun 3, 2006

### tuna_wasabi

I'll post these in two messages to make this easier. First problem:
I'm pretty stuck on this one. I've tried juggling all the kinematics equations but the best I can do is solve x=v*cos(theta)*t for t and drop it into y=(1/2)gt^2 + v*sin(theta)*t. This yields the standard Range equation, but I can't solve that for theta. I'm just not sure how to derive the right equation.

2. Jun 3, 2006

### tuna_wasabi

Never mind the second problem, I figured it out as I was typing it. Thanks in advance.

3. Jun 3, 2006

### dimensionless

First off, gravity counters the vertical motion of the cannon ball. Your y-equation should thus be y(t)=-(1/2)gt^2+v*sin(theta)*t

To find theta you will need to determine how long it will take the ball to x(t)=2000m. Once you have found 't' you can then substitute it into your y-equation and solve for theta.

4. Jun 3, 2006

### tuna_wasabi

Yes, but to know how long it takes the ball to travel 2000m along the ground I need to know the x-component of the initial velocity. But I can't get the x-component without theta, which is what I'm trying to find.

5. Jun 3, 2006

### finchie_88

You could always combine the x and y component equations (eliminating t in the process), and solve the resulting equation for the angle, it would probably involve more work, but it would work best overall.

6. Jun 3, 2006

### tuna_wasabi

As stated above, I've tried that. Solving x=vi*cos(theta)t for t and putting into y=(1/2)gt^2 + v*sin(theta)*t yields the standard projectile motion range equation, which is [deep breath]:

y = tan(theta)*x - [(g*x^2)/(2*vi^2*cos^2(theta))]

If you can solve that for theta, I will bow to your awesome trig skills.

7. Jun 6, 2006

### siddharth

Try using

$$\frac{1}{\cos^2 \theta} = sec^2 \theta$$

$$sec^2\theta = 1 + \tan^2 \theta$$

and solve for $\tan \theta$.

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