Needed impulse to move an object to a point

Topics: Developer Forum, User Forum
Sep 23, 2009 at 4:51 PM
Edited Sep 23, 2009 at 4:54 PM

Hi there, more than an engin question, is a physics question. This is what I'm doing:

I have a body with mass M, in an environment with gravity G. I need to know how much impulse I need to make the object "jump" to a certain point in the world.

First for the height (the Y factor) I know that the impulse is the delta of the momentum, and I know that:

D = Vi + 1/2 * A *T^2

(D = Distance traveled, Vi = initial velocity, A = Acceleration, in this case the gravity in negative, T = time)

And the velocity is

Vf = Vi + A*T

(Vf = final velocity)

I want, of course my final velocity to reach 0 when it arrives to destination, so making Vf = 0, I get that Vi = -A*T, and as the gravity is the acceleration (in negative), I have Vi = G*T. Replacing this I have in the first equation:

D = G*T - 1/2 * G * T^2

1/2*G*T^2 - G*T + D = 0

I know D, I know G, I need to know T. Using Baskara formula I have that:

T = (G +- SQRT((-G)^2 - 4 * 1/2*G * D))/2*1/2*G

Now I have the time needed to jump to that height. Returning to the first equation (D = Vi + 1/2 * A *T^2), if I multiply everything by the body's mass I get:

D*M = Vi*M + 1/2 * A * T^2 *M

The impulse is the difference of the momentum (Velocity * Mass), so replacing Vi*M by impulse (I), I get:

D*M = I + 1/2 * A * T^2*M

Then I have the needed impulse I.

But that isn't working for me. What I'm I doing wrong? Please help!

 

Cheers

Sep 28, 2009 at 11:33 AM

You should try a simpler method for finding the impulse.

p = M * v

momentum is mass times velocity. Since an impulse is an instantaneous change in momentum the only thing you need to find out is the initial velocity for the jump (Vi), then multiply it by the object's mass(M).

Now to find that we use one of the equations you already got

V = Vi - G * t

and find the time it takes for V to equal zero.

0 = Vi - G * t

G * t = Vi

t = Vi / G

Now that we have the time we substitute it into our equation for distance.

d = Vi * t - G * t^2 / 2

and we get

d = Vi * Vi / G - G * Vi^2 / (2 * G^2)    note that I also substitute t^2 for Vi^2 / G^2, you forgot to do that on your post, probably what got you the wrong answer

and we simplify to

d = [Vi^2 / G] - 1/2 [Vi^2 / G]             note the parts between the brackets

and then to

d = 1/2 Vi^2 / G

now we can solve for the initial velocity of the jump (Vi)

2 * d = Vi^2 / G

Vi^2 = 2 * d * G

Vi = (2 * d * G)^(1/2)                           note x^(1/2) = sqrt(x)

multiply that by your objects mass and you should get the impulse you need

I = M * (2 * d * G)^(1/2)

 

Si alguien presenta gripa debe ser tratado por un médico antes de 48 horas.

 

·         La vacuna contra la influenza no provee ninguna protección contra la gripa porcina.

Sep 28, 2009 at 12:32 PM

@vulgivagus You're absolutely right. Amazing, now it works like a charm. Thank you very much!