Classical Mechanics

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_PJ_(Posted 2016) [#1]
Can anyone see what's wrong with this?

In a classical, non-relativistic system

An idealised, frictionless point mass of "m" = 1 kg

Launched vertically (only dimension required x1) from x(t=0)=1 metre above the ground.

The direction is directly opposite the gravitational field imparting a force which provides acceleration "g" metres per second per second.

The projectile is launched with a SPEED of 5 metres per second

Potential Energy is maximal at time tA, when the object is at height of "A" metres vertically above ground along x1 axis.

The speed at which the projectile impacts the ground (assuming whatever fired it has been removed) at time tB is "B" metres per second at which point, it is assumed potential energy = 0

This is summarised:

TIME T=0

X = 1
X' = 5
X''= g

Non-Relativistic, Classical Energy:
El = 12.5
Ev = g


TIME T=tA

X = A
X' = 0
X''= g

Non-Relativistic, Classical Energy:
El = 0
Ev = g*A

TIME T=tB

X = 0
X' = B
X''= g

Non-Relativistic, Classical Energy:
El = ½*B^2
Ev = 0

Assuming no thermal energy loss, all energy is conserved therefore

E0 = EA = EB = 12.5+g = g*A = ½*B^2

It follows:
A = (½B^2)/g
B = Sqr(2*g*A)

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There seems to be issues with the signs.


Floyd(Posted 2016) [#2]
At T=0 you have velocity is 5 and both acceleration and potential energy equal g.
But potential energy is positive while acceleration is negative since it decreases a positive velocity.

Anyway, velocity is positive moving up and negative moving down. Energy can't be negative since it is based on velocity squared.


Matty(Posted 2016) [#3]
.