I have a question about this classical mechanics application.(adsbygoogle = window.adsbygoogle || []).push({});

Given is a point with mass m and speed v0. It gets shot from the ground under an angle alpha.

Wanted is the path of this projectile.

This is a two dimensional example so I need to find the motion equation.

I know that m*x(double dot) = Fx = 0

and m*y(double dot) = Fy = -m*g

so x(double dot) = 0

and y(double dot) = -g

The initial speed is given by :

v0 = v0*cos(alpha)*1x + v0*sin(alpha)*1y (vectorial notation)

Then they say that the first integration regarded to t gives:

x(dot) = constant = x(dot)(0) = vo*cos(alpha) and

y(dot)(t) = -g*t + constant = -g*t + v0*sin(alpha)

I don't see how they come to this result. Can someone explain this?

Further they say that the second integration regarded to t gives:

x(t) = (v0*cos(alpha))*t

y(t) = -1/2*g*t² + (v0*sin(alpha))*t

Then they say that the path of the object is in the x,y plane with equation y = F(x)

Now they ask me to find that F(x).

They say you can do it by elimination of the variable t (t= x/(v0*cos(alpha))) Can anyone explaine me how they find this value for t and how this elimination procedure works?

Many thanks in advance!

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# Shot parabola

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