Help with optimization question

Oct 2019
Hi could i please get help with this question ?
A grapefruit is tossed straight up with an initial velocity of
ft/sec. The grapefruit is
feet above the ground when it is released. Its height at time
is given by​

How high does it go before returning to the ground?
Dec 2014
position as a function of time

$y=y_0 + v_0 t - \dfrac{1}{2}gt^2$

$\dfrac{dy}{dt} = v_0 - gt$

max height occurs when $\dfrac{dy}{dt} =0 \implies t = \dfrac{v_0}{g}$

substitute $\dfrac{v_0}{g}$ for time, $t$, in the position equation to determine maximum height as a function of initial position, $y_0$, initial velocity, $v_0$, and the magnitude of acceleration due to gravity, $g$.

BTW ... some links in your post are not showing, at least for me they’re not.
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