What is the graph obtained when a body is thrown in upward direction with velocity on y-axis and position of body on x-axis? Also,can you please explain how it is obtained?

Aug 12, 2018

Parabola

Explanation:

When a body is thrown upward,it moves upward first because of the imparted velocity to it,but gravity opposes this motion and it comes to rest after attaining certain height and then starts falling back to the ground due to gravity!

Now,when it is going up, it is having negative acceleration upwards better say as deceleration upwards and that is constant!

So,we can write,

$\frac{\mathrm{dv}}{\mathrm{dt}} = - g$

$\implies \frac{\mathrm{dv}}{\mathrm{ds}} \frac{\mathrm{ds}}{\mathrm{dt}} = - g$

$\implies v \frac{\mathrm{dv}}{\mathrm{ds}} = - g$

$\implies v \mathrm{dv} = - g \mathrm{ds}$

$\implies {\int}_{u}^{v} v \mathrm{dv} = - g {\int}_{0}^{h} \mathrm{ds}$ (let,it was thrown upward with velocity $u$)

so,$\frac{{v}^{2} - {u}^{2}}{2} = - g h$

or, ${v}^{2} = {u}^{2} - 2 g h$

So,this is the equation which establishes the relationship between velocity and position in this case.

From the equation,we can see ${v}^{2} \propto h$,which is an equation of parabola!

And,it looks like as follows-
graph{y^2=400-19.6x [-80, 80, -40, 40]}

Consider the part in $1$ st Quadrant only.

Veocity is shown along $Y$ axis and position along $X$ axis. Here, I took $u = 20 m {s}^{-} 1$ to show you and $g = 9.8 m {s}^{-} 2$

So,you can see if you go from left to right of the curve i.e with increasing height,velocity starts from $20 m {s}^{-} 1$ to $0$ which happens when it reaches at its maximum height!

Then it falls down which you can get by back tracing the curve i.e velocity increases from $0$ to $20 m {s}^{-} 1$ when it reaches ground that is at point $0$ along $X$ axis.