Recent content by ChromoZoneX

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    Inverse Function problem involving e^x

    Thank you very much. I understood the question and answered it PS: Sorry for the late reply.
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    Inverse Function problem involving e^x

    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?
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    Finding Initial Velocity for Projectile Launch at Angle Theta

    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.
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    Finding Initial Velocity for Projectile Launch at Angle Theta

    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?
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    Finding Initial Velocity for Projectile Launch at Angle Theta

    Your expression for 't' is equivalent. I worked that out. :D
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    Finding Initial Velocity for Projectile Launch at Angle Theta

    It is stretched before launch so that it can launch itself with the energy stored in it.
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    Finding Initial Velocity for Projectile Launch at Angle Theta

    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...
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    Finding Initial Velocity for Projectile Launch at Angle Theta

    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.
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    Finding Initial Velocity for Projectile Launch at Angle Theta

    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)?
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    Finding Initial Velocity for Projectile Launch at Angle Theta

    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)
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    Finding Initial Velocity for Projectile Launch at Angle Theta

    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)
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    Finding Initial Velocity for Projectile Launch at Angle Theta

    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...
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    Finding Initial Velocity for Projectile Launch at Angle Theta

    Yes that is correct. Given height, distance and angle of launch, find velocity. PS: That is an awesome drawing!
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    Finding Initial Velocity for Projectile Launch at Angle Theta

    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...
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    Finding Initial Velocity for Projectile Launch at Angle Theta

    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.