# Determine the impulse required for a projectile to reach to specific height with a variable initial velocity [closed]

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.

• @PM2Ring My apologizes I forgot to include the $\sqrt{}$ – grahamcracker1234 Jun 2 '20 at 17:32

## 1 Answer

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,

$$I(t)=-mgt$$

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

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