Recent content by ChromoZoneX

Thank you very much. I understood the question and answered it PS: Sorry for the late reply.

Homework Statement Let g(x) = (e^x - e^-x)/2. Find g^-1(x) and show (by manual computation) that g(g^-1(x)) = x. Homework Equations g(x) = (e^x - e^-x)/2 The Attempt at a Solution I get the inverse = ln[ (2x + sqrt(4x^2 + 4) ) / 2 ] How do I proceed?

As far as I've been told, for our theoretical calculations, we're supposed to assume that all of the spring's elastic energy gets converted to kinetic energy. That is how I ended up with this, mv^2 = kx^2 And finding 'x' as a result of all that other stuff in the thread.

The spring is the projectile itself. It must be stretched at a given angle so that it can hit a target that is some horizontal distance away and some vertical distance high. I need to find how much it is stretched. What is the solution you propose?

Your expression for 't' is equivalent. I worked that out. :D

It is stretched before launch so that it can launch itself with the energy stored in it.

Now, I'm confused. Let me explain the whole scenario. I have to make a spring launcher that will fire a stretched spring. I will be given a target that is 'x' meters away and 'y' meters high w.r.t launch point. I will also be given an angle of launch theta. My solution (as of now), x...

All I did is switch it to , y=V Sin( \theta ) t + \frac{g}{2} t^2 x=V Cos( \theta )t and then divided y/x to get tanθ and hence the equation.

Also, when we talk about 'x', is it the total distance traveled by the projectile or is it just half the distance (that is the horizontal distance covered when it has reached its maximum height)?

x = vcos\vartheta.t t = x/(vcos\vartheta) tan\vartheta=(y-0.5gt2)/x Isolating 't', t2 = (x tan\vartheta - y)/-0.5g t = \sqrt{}((x tan\vartheta - y)/-0.5g)

OK this is my equation for 't', t=\sqrt{}((xtan\vartheta - y)/(-0.5g)) Can I substitute this for 't' in the equation, v = x/(cos\varthetat)

I have the angle theta, the distance the target is away and the height of the target. I don't have the time. I need to find the initial velocity of the projectile. From your equations, I get the "v sin" and "v cos" equations for x and y. Shouldn't the 'x' eqn use v cos(theta) and 'y' be v...

Yes that is correct. Given height, distance and angle of launch, find velocity. PS: That is an awesome drawing!

Yeah, I mean maximum height achieved by the projectile. It's a spring that needs to be stretched 'x' so that it attains a certain initial velocity 'v' so that it hits a target 'd' meters away and 'h' meters high from the same level of launch. I know that x = √(mv2/k) I know the value of...

How to find the initial velocity required for a projectile to be launched at an angle theta, with a distance 'd' and height 'h'. An expression is required. I am unable to find one using all three. Thanks in advance.