58. To find the “launch” velocity of the rock, we apply Eq. 2-11 to the maximum height (where the speed is momentarily zero) ()()20009.8 m/s2.5 svvgtv=−¡=−so that v0= 24.5 m/s (with +yup). Now we use Eq. 2-15 to find the height of the tower
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