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Projectile Motion's relationship with Kinetic Energy and Potential Energyby Whakataku Tags: kinetic energy, potential energy, projectile motion, range 

#1
Oct711, 01:01 AM

P: 12

Imagine a ramp setup on top of a tall table. The height Δy is measured. To find the initial velocity at the instant the ball leaves the ramp, I set up the kinetic energy and potential energy equal to each other to find the initial velocity of the x component.
PE = KE m*g*(hr) = 0.5*m*v^2 where hr is the height of the ramp and v is initial velocity (xcomponent) solving for vx (xcomponent velocity), I got: vx = √(2*g*hr) To get the time for the object's time in flight: y'y= vy + 0.5gt^2 Δy= vy + 0.5gt^2, where Δy is the height from the ground to the ramp. since θ= 0° I found t to be: t = √{ (2*∆y)/g } Now my question is how do I find the range of this object? I started out with Δx = vx*t ; where vx is the initial xcomponent velocity.... is that even right? I'm hesitant to use it because written as Δx/t, it looks like an average velocity equation. Furthermore, in Wikipedia I saw the equation d= {v*cos([itex]\Theta[/itex])}/g * [v*sin([itex]\Theta[/itex]) + sqrt(v*sin([itex]\Theta[/itex])^2+2g*y)] http://en.wikipedia.org/wiki/Range_of_a_projectile under uneven ground but the problem is I don't have final velocity................ or can I calculate the final velocity with the givens........ if so how?? Could anyone please nudge me in the right direction to find Δx? thanks. 



#2
Oct711, 01:19 PM

P: 12

I think I solved it........... duh.
in the distance formula d = v(initial)*t + 1/2*a*t^2. v(initial) and time is already attained and the the a acceleration is 9.81m/s^2 Correct me if I'm wrong. 


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