How do you find the angle and the initial velocity of a trajectory

>>8293146
>pokemans memes on sci

>>8293146
you might be able to derive something from the range equation. im not that great at physics though.

File: dontwanttolivehere.jpg (7KB, 215x120px) Image search: [Google]

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draw a fucking parabola

some combination of kinematic equations. you can find the time it was in the air only using it's max height, from there you can find it's horizontal velocity, so there's a start.

>>8293146
Refer to your kinematic equations for constant acceleration. If you know the max height, you are able to find the initial velocity in the y-direction. You can then use that initial velocity to find the time it took to get up to the max height. Once you find the time it took to travel to the max height, finding the max horizontal distance is straight foward

>>8293146

You can't. You need to specify the parameters of the object in addition to specifying any additional frictional sources that might alter the dynamical path.

If you're in an introductory physics class, then you can most likely assume that the whole motion is conservative and a parabola in which case this is just algebra I.

File: diff.png (17KB, 541x286px) Image search: [Google]

17KB, 541x286px

>>8293177
Thank you, that worked but I am not sure why

What's the difference between these?

>>8293208
that's stupid...
They're the max over all the theta_0's
max altitude is at theta_0=90 deg
max range is at theta_0=45 deg

>>8293208
It works because the kinematic equations with constant acceleration are dependent on time. Both the y-displacement and x-displacement go off the same time. So if it takes 10 seconds for an object to travel to max height and fall back down, it will also take 10 seconds for the same object to travel to its maximum horizontal distance.

To see where your equations in the picture come from, recall the kinematic equation [math]v^{2} = v^{2}_{0}+2a\Delta r[/math]. If we solve for the displacement r, we get
[math]\Delta r = \frac{v^{2}-v_{0}^{2}}{2a}[/math]
Note the velocities you have in the picture, which are represented in terms of the angle. If you substitute those into the above equation and take the acceleration to be g, you will get back those in the pic.

For the max ranges, all we need to see is when [math]\sin^{2}\theta[/math] and [math]\sin(2\theta)[/math] are maximized. Try to figure these out for yourself.

>>8293234
I don't get it

height and range are both 100m

I have a system that's

2*9.8*100=v^2 * (sinθ)^2

9.8*100=v^2 * (sin2θ)

Why?

>>8293272
One gives you displacement of the object along the y direction, or altitude as they call it, and one gives you the displacement along the x axis, or horizontal range. The object has both an x and y coordinate to denote its position in space. If it is travelling along a path, that position can have a combination of x and y coordinates. Thats why there are two separate equations. They are related by the angle at which your object is projected at, which should be obvious if you draw out the velocity vector for your object.

>>8293146

Seriously its simple. Take the time it takes for object to fall from max height. Use time to calculate what speed is needed to cross horizontal distance in that time.

Woooow was it that hard?

>>8293483

Oh and don't forget to take it times too but that should be obvious

>>8293146
This is fucking high-school applied maths. Give us the numbers anyways.