ID	ef3322bd-39b9-433f-8f5e-52458722320c
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2026-02-12

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cole physics problem 1: colliding balls

ID	27dc2d74-cb11-490c-bb5c-be0a657899a0
DEERTOPIAVISIBILITY	public

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set up position functions. we'll call the object being thrown $A$ , and the object being dropped $B$ . $\begin{matrix} s_{A} (t) & = v_{0} t + \frac{1}{2} g t^{2} \\ s_{B} (t) & = H + \frac{1}{2} g t^{2} \end{matrix}$

part I

set the position functions equal to each other and solve for $t$ to find out when they collide. $\begin{matrix} s_{A} (t) & = s_{B} (t) \\ v_{0} t + \frac{1}{2} g t^{2} & = H + \frac{1}{2} g t^{2} \\ v_{0} t & = H \\ t & = \frac{H}{v_{0}} . \end{matrix}$ from now heretothitherforth let $t_{collision} ≔ \frac{H}{v_{0}}$ .

part II

first we'll find the time of ball $A$ 's apex (a critical point) by solving $s_{A}^{'} (t) = 0.$ for $t$ . chop chop sydney durr durr $s_{A}^{'} (t) = v_{0} + g t$ and blahba $t_{apex} = - \frac{v_{0}}{g} .$

awesome now if we want the balls to collide at $A$ 's apex we just set $t_{apex} = t_{collision}$ and solve for $H$ $\begin{matrix} t_{apex} & = t_{collision} \\ - \frac{v_{0}}{g} & = \frac{H}{v_{0}} \\ - \frac{v_{0}^{2}}{g} & = H . \end{matrix}$