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Projectile Motion's relationship with Kinetic Energy and Potential Energy

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Whakataku
#1
Oct7-11, 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 (x-component)

solving for vx (x-component 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 x-component 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.
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Whakataku
#2
Oct7-11, 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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