# Chapter 2 - Exercises and Problems - Page 33: 97

a) $v_0>\sqrt{.5hg}$ b) $y = h -\frac{gh^2}{2v_0^2}$

#### Work Step by Step

a) We know the following equation for the value of y, recalling that $a=g$ in this case since the only acceleration is due to gravity: $y = v_0t+\frac{1}{2}gt^2$ We also know: $y=h-\frac{1}{2}gt^2$ Setting these equal gives: $v_0t+\frac{1}{2}gt^2=h-\frac{1}{2}gt^2$ $v_0t=h \\ v_0=\frac{h}{t}$ We know that the value of t is $t = \sqrt{\frac{2h}{g}}$, so we obtain: $v_0>\frac{h}{\sqrt{\frac{2h}{g}}}$ $v_0>\sqrt{.5hg}$ b) We use the second equation, which is $y=h_0-\frac{1}{2}gt^2$. Using substitution, we find: $y = h_0 -\frac{gh_0^2}{2v_0^2}$

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