# Projectiles physics help

1. Dec 5, 2006

### m2287

1. The problem statement, all variables and given/known data

the initial speed of a projectile is v0. what angle of projection theta makes its horizontal range R a maximum? what is the maximm horizontal range?

2. Relevant equations

3. The attempt at a solution

i think that for a maximum R, theta is 45degres

to work out the maximum horizontal range im guessing i have to resolve horizontaly or something and the furthest i get is v0(x)=v0cos(theta)

if anyone could give any help i would be extremely gratefull

2. Dec 5, 2006

### neutrino

Can you derive/Do you know the expression for the horizontal range of a projectile?

3. Dec 5, 2006

You are right, the maximum angle is 45 degress, but now you have to proove that. I suggest you start with writing down all relevant equations.

4. Dec 5, 2006

### m2287

i can do differentiation but i dont know the horizontal expression of a projectile?

so you mean the s u v a t equations? like s = ut + 0.5 at^2

5. Dec 5, 2006

Hint: if you can find the expression for the trajectory, you can easily get the expression for the horizontal range (dependent on the angle) of the projectile from the fact that it equals twice the maximum height of the projectile. I can't think of an easier way right now, so maybe you get a better suggestion, but if you ask me, do the following:

1. Find the equation of the trajectory
2. Find the maximal height
3. Multiply the height by 2 to get the range of the projectile
4. Differentiate the expression for the range with respect to the angle, and set it equal to zero.

6. Dec 5, 2006

Please ignore my previous post. I found the easiest way.

So, you know the equation x(t). All you have to do is find the time t at which the projectile falls to the ground. Plug that time into the equation and differentiate with respect to the angle. Set it equal to zero, and your problem should be solved. Once again, sorry if I confused you with the upper post.

7. Dec 5, 2006

### m2287

ok so i get:

x(t) = v0cos(theta)t + 0.5 * -10t^2

x(t) = v0cos(theta)t -5t^2

d(x)/d(t) = -v0sin(theta)t - 10t

d(x)/d(t) = -v0*0.525t - 10t

@ d(x)/d(t) = 0 ----> 0 = -v0*0.525t - 10t

v0 = -10t/0.525t

v0 = -19

surely this should be positive!?

8. Dec 5, 2006

You're mixing up equations for x(t) and y(t). There is no acceleration in the x-direction, so you have $$x(t) = v_{0}\cos \theta \cdot t$$. Now find the time it takes for the projectile to fall to the ground (i.e. solve the equation y(t) = 0) and plug it into the equation x(t).

9. Dec 5, 2006

### m2287

so for y(t) i get

y(t) = 0.5 * -10t^2

0 = -5t^2

0 = 0

surely putting y(t) as 0 gives you y when it has traveled no distance?

:S!

10. Dec 5, 2006

### neutrino

Isn't y(t) = v_0*sin(theta)t - 0.5gt^2?

11. Dec 5, 2006

It sure does, bot not if you use the wrong equation. Use the equation suggested by neutrino.

12. Dec 5, 2006

### m2287

ok so i get:

@ y(t) = 0, t = 0.17v0

putting that into x(t) = v0cos45t

x(t) = v0 * 0.525* 0.17v0

x(t) = 0.0893v0^2

does this look right?

13. Dec 5, 2006

$$y(t) = v_{0}\sin \theta \cdot t - \frac{1}{2}gt^2 = 0 \Rightarrow$$
$$t=\frac{2v_{0}\sin \theta}{g}$$.
After plugging the time into $$x(t) = v_{0}\cos \theta \cdot t$$, you have:
$$x = \frac{2 v_{0}^2}{g}\sin \theta \cos \theta =$$
$$\frac{v_{0}^2}{g}\sin (2\theta)$$. (after using sin(2theta) = 2 sin(theta)cos(theta). )

Now differentiate the expression with respect to $$\theta$$ and set it equal to zero.

14. Dec 5, 2006

### m2287

thanks alot for your help, i think i have it now :D!