Thread: shot parabola
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Integral is offline
Nov13-03, 09:19 AM
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Originally posted by Flukkie
I have a question about this classical mechanics application.

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(dubble dot) = Fx = 0
and m*y(dubble dot) = Fy = -m*g

so x(dubble dot) = 0
and y(dubble 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?

I wish we had the dot notation! You will find that, even if it does not offer everything, this site offers much more then most in the way of notation.

To your question, let me use a prime (') in place of the dot
We have
x'(t) = C
Note that I added the implied variable, now we need to evaluate the constant for all time, we have the initial condition of


But since x'(t)=C we must have C=v0Cos(α)
repeat this logic to get evaluate your constant of integration for the y equation.

Further they say that the second integration regarded to t gives:
x(t) = (v0*cos(alpha))*t
y(t) = -1/2*g*tē + (v0sin(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!
Simply solve your equation for x(t) for t, then take the resulting function of x and subsitiute it into your y(t) equation.

In the future you may wish to post such questions to our Homework help fourm.