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I’m having a hard time trying to figure this question out.

With what initial velocity must an object be thrown upward (from ground level) to reach the top of the Washington Monument, which has a height of approximately 555 feet? (Use -32ft/sec^2 as acceleration due to gravity)

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  • $\begingroup$ Well, use $h(t) = h_0 + v_0t + \frac{1}{2}gt^{2}$, find the maximum depending on $v_0$ and $g$, and solve for $v_0$. $\endgroup$ Commented Feb 24, 2021 at 19:37

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Let $ H $ be the height of the monument.

The object reaches the top, when its velocity $ V_1=0 $.

Let $ V_0 $ be the initial velocity.

You just need and you should, know that

$$V_1^2-V_0^2 = -2g(H-0)= -V_0^2$$

thus

$$\boxed{V_0=\sqrt{2gH}}$$

with $$g=32 \text{ and } H=555$$

Remark :

Other way to get the formula above. It attains the top, after a time $ T $ such that $$H=-\frac 12 gT^2+V_0T$$

with

$$V_1=-gT+V_0=0$$ So

$$T=\frac{V_0}{g}$$ and $$H=-\frac 12 g(\frac{V_0}{g})^2+V_0.\frac{V_0}{g}$$ $$=\frac{V_0^2}{2g}$$

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