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A mountain climber jumps a crevasse of width WW by leaping horizontally with speed v0v_0. If the height difference between the two sides of the crevasse is hh, what is the minimum value of v0v_0 for the climber to land safely on the other side?

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We can start solution with equations for projectile motion:

h=12gt2t=2hgd=W=v0xtv0=Wtv0=Wg2h\begin{align*} h&=\frac{1}{2}gt^2 \tag{vertical motion} \\ t&=\sqrt{\frac{2h}{g}} \tag{express $t$} \\ d&=W=v_{0x} t \tag{horizontal motion} \\ v_0&=\frac{W}{t} \tag{substitute $v_{0x}$} \\ v_0&=\boxed{W \sqrt{\frac{g}{2h}}} \tag{substitute $t$} \end{align*}

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