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# Stupid basic physics q

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If you throw an object straight up at any given velocity (not accounting for drag etc) wont it be at the same (negative) velocity at the same y point on the way back down? Thanks in advance. Cheers.

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Yes. I don''t know any way to go about showing it other than just calculating it up i guess, but yes, it''ll be at an equal negative velocity at the same y coord.

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possibly, try using the equation

V = Vo +aT

where
V=final velocity
Vo=initial velocity
a=acceleration (in this case gravity), and
T= time

now, its late and im tired so i will rely on the other gamedevrs to check my work. lets say Vo = 20 m/s, a = -9.8m/s^2, we now have two unknowns, V and T, we are checking to see if the final velocity and initial are the same magnitude but opposite direction so...

up is positive

-20 = 20 - 9.8T
-9.8T = -40
T = ~4.1s

that means that the initial and final velocties will be equal and opposite if the object is in the air for 4.1seconds

edit: this could be shown much easier using a velocity-time graph but i dont feel like drawing
edit: damn im really tired, im reading my above post and im not sure that i proved that its the same, i think i just solve for the time, which doesnt prove anything, well, ive still got my dignity, and a 90% in physics, if no one else is able to give u a more concrete answer i will try again tomorrow

,Matt

-= kill one your a murderer, kill thousands your a conquerer =-

[edited by - samosa on March 3, 2003 10:51:22 PM]

[edited by - samosa on March 3, 2003 10:53:39 PM]

[edited by - samosa on March 3, 2003 10:54:21 PM]

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You can show that its true analytically with conservation of energy. If you neglect air resistance, the total sum of the potential and kinetic energy of some object will be constant during the motion of the object. The potential energy is a function of the distance from the earth which is equivalent to the y-coordinate. Since the potential energy at any height is constant, and the total energy is constant, the kinetic energy is constant. For a particular mass, only one kinetic energy exists for a particular velocity, so the velocity is the same at any point on the way up as it is on the way down. Just in case you cared.

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Yup. And no I dont want to do integrals to prove it right now. I''ve got a math test tomorrow, damnit...

stupid Math 125 ^_^

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conservation of energy of the projectile means that

0.5*m*v^2 + m*g*h1 = 0.5*m*u^2 + m*g*h2

cancel the mass

0.5v^2 + g*h2 = 0.5*u^2 +g*h1

we want the velocity at the time when h2=h1 so we can cancel it off

0.5v^2 = 0.5u^2

v^2=u^2

taking the square root of both sides gives us

+/- v = +/- u

so the magnitude of the velocities will be equal at a given height.. but the direction of the velocity wont be

so on the way up.. v is positive .. on the way down u=-v
(because velocity has direction)
the speeds however are equal

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