Maximizing Horizontal Range: Projectile Speed and Angle | Physics Homework Help

[m2287]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

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

to work out the maximum horizontal range I am 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

 

[neutrino]

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

 

[radou]

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.

 

[m2287]

i can do differentiation but i don't know the horizontal expression of a projectile?

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

 

[radou]

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.

 

[radou]

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.

 

[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!?

 

[radou]

ok so i get:

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

You're mixing up equations for x(t) and y(t). There is no acceleration in the x-direction, so you have [tex]x(t) = v_{0}\cos \theta \cdot t[/tex]. 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).

 

[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!

 

[neutrino]

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

 

[radou]

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!

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

 

[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?

 

[radou]

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?

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

Now differentiate the expression with respect to [tex]\theta[/tex] and set it equal to zero.

 

[m2287]

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