# A body is thrown into the air with an initial velocity of $v_0$ ft/sec. What initial velocity is required to double the maximum height?

I'll post the question here and then show my workings out.

A body is thrown into the air with an initial velocity of $$v_0$$ ft/sec. What initial velocity is required to double the maximum height previously attained?

First I write the equation for velocity in feet.

$$v=-32t+v_0$$

Integrate to get s.

$$s=-16t^2 + v_0t$$

Calculate the maximum.

when $$v=0$$, $$\frac{v_0}{32}=t$$

Twice the distance is $$2s=-32t^2+2v_0 t$$

Substituting $$\frac{v_0}{32}=t$$ into this equation gives $$2s=\frac{-v_0^2}{32}+\frac{v_0}{4}$$

I then differentiate $$2s$$, now as $$s_2$$ to get the velocity required in terms of initial velocity.

$$v_2=\frac{4-v_0}{16}$$ $$v_2=\frac{2-\sqrt(v_0)}{4}$$

This is not the correct answer, suggestions would be greatly appreciated!

• You're missing a constant of integration at the first $s$ step, although I'm not sure if that matters
– Zim
Commented Oct 9, 2020 at 15:35
• Is this formula not known?$$v^2-{v_0}^2=2a\Delta x$$ Commented Oct 9, 2020 at 15:41
• "... Substituting $\frac{v_0}{32}=t$ into this equation gives..." Of course the new $t$ is different than the old $t$. Nobody can guarantee that the body will achieve the maximum height in the same time in both cases.
– user700480
Commented Oct 9, 2020 at 15:49
• I am not familiar with $v^2−v_0^2=2a\delta x$ Commented Oct 9, 2020 at 19:01
• Thank you Stinking Bishop, that makes sense. Commented Oct 9, 2020 at 19:01

## 1 Answer

Recall that

$${v_f}^2-{v_i}^2=2a\Delta x$$

where $$v_i$$ and $$v_f$$ denote initial and final velocities, respectively; $$a$$ is acceleration; and $$\Delta x$$ is the net displacement.

At its maximum height, the object has $$0$$ vertical velocity, so if it is thrown with initial velocity $$v_0$$ and is in freefall, then

$$0^2-{v_0}^2=-2g\Delta x\implies \Delta x=\dfrac{{v_0}^2}{2g}$$

where $$g$$ is the magnitude of the acceleration due to gravity.

If one wants the object to reach a maximum height twice as high, that would require scaling up $$v_0$$ by a factor of $$\sqrt2$$, since

$$\dfrac{(\sqrt2\,v_0)^2}{2g}=\dfrac{{v_0}^2}g=2\Delta x$$