Classical physics (phase diagram

[dirac1902]

A ball of mass m rests at times t less than 0, at height h above the ground. at time t=0 the ball is released and fell on the ground after hitting the ground, the ball rests there.

a) x represents distance from the ground, drive and expression for the phase diagram when the ball moving toward the ground. and draw it and find where v and x meet the axis's.
The attempt at a solution

1/2(mv^2)= mg(h-x)

v=sqrt(2g(h-x)) #

the graph will be a curve (square root) the start will be when x=h and then when it hits the ground will have max v so the position will be zero and v is max.

Thanks

 

[cepheid]

A ball of mass m rests at times t less than 0, at height h above the ground. at time t=0 the ball is released and fell on the ground after hitting the ground, the ball rests there.

a) x represents distance from the ground, drive and expression for the phase diagram when the ball moving toward the ground. and draw it and find where v and x meet the axis's.



The attempt at a solution

1/2(mv^2)= mg(h-x)

v=sqrt(2g(h-x)) #

the graph will be a curve (square root) the start will be when x=h and then when it hits the ground will have max v so the position will be zero and v is max.

Thanks

It all seems fine to me. The question asks you for the point at which the curve meets the two axes (plural of axis by the way), in other words it asks for the the x-intercept and v-intercept. So you might want to state more explicitly that:

- the x-intercept (when v = 0) occurs at x = h

- the v-intercept (when x = 0) occurs at y = √(2gh)

 

[dirac1902]

It all seems fine to me. The question asks you for the point at which the curve meets the two axes (plural of axis by the way), in other words it asks for the the x-intercept and v-intercept. So you might want to state more explicitly that:

- the x-intercept (when v = 0) occurs at x = h

- the v-intercept (when x = 0) occurs at y = √(2gh)

but how the curve will be. I mean will it open up or down?

Thanks for your help :)