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.(adsbygoogle = window.adsbygoogle || []).push({});

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" [Broken]

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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# Projectile Motion's relationship with Kinetic Energy and Potential Energy

## if two different masses have equal kinetic energy, they must not have equal momentum.

Poll closed Oct 14, 2011.

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