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Assume we have a projectile which we want to shoot straight up into the air, such that we are only working with the y-component, what formula can be used to determine the impulse required for this projectile to reach a certain height?

I was able to derive the following formula assuming no initial velocity:

$$F\Delta t = m\sqrt{2gh}$$

with $F$ being the force applied to the projectile, $\Delta t$ being amount of time the force is applied, $g$ being the acceleration due to gravity, $m$ being the mass of the projectile, and $h$ being the height that the projectile must reach.

Is there way to derive this formula to include an initial velocity? My attempts have all failed.

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  • $\begingroup$ @PM2Ring My apologizes I forgot to include the $\sqrt{}$ $\endgroup$ – grahamcracker1234 Jun 2 at 17:32
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The impule is $I=\Delta p= m \Delta v$. The equations of motion for the object will be:

$$ h(t)=h_0+v_0 t-\dfrac{1}{2}gt^2$$ $$ v(t)=v_0 -gt.$$

But remember you can combine both to get

$$v_f^2-v_0^2=2g\Delta h.$$

So that you can have various equations for the impulse, depending on what you want to do,

$$I(t)=-mgt$$

$$I(h)=m(\sqrt{v_0^2+2g\Delta h}-v_0).$$

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  • $\begingroup$ I may be incorrect, but should the $v_0^2$ actually be $v_f^2$? $\endgroup$ – grahamcracker1234 Jun 8 at 20:57
  • $\begingroup$ In which formula? C: $\endgroup$ – vin92 Jun 9 at 8:44
  • $\begingroup$ I believe the last formula should be $I(h)=m(\sqrt{v_f^2+2g\Delta h}-v_0).$ $\endgroup$ – grahamcracker1234 Jun 9 at 18:00
  • $\begingroup$ But if you solve for $v_f$ in the equation $v_f^2-v_0^2=2g\Delta h$, you get $v_f=\sqrt{v_0^2+2g\Delta h}$ right? $\endgroup$ – vin92 Jun 9 at 23:50

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