A ball thrown into the air from a building 60 mtrs high,travels along a path given by h(x)= 60+x-x2/60,
where x is the horizontal distance from the building and h(x) is the height of the ball.What is the maximum height the ball will reach?

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{h}\left(\mathrm{x}\right)=60+\mathrm{x}-\frac{{\mathrm{x}}^2}{60} \\ \mathrm{Differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x}, \mathrm{we} \mathrm{get}: \\ \mathrm{h}\text{'}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left\{60+\mathrm{x}-\frac{{\mathrm{x}}^2}{60}\right\} \\ \Rightarrow \mathrm{h}\text{'}\left(\mathrm{x}\right)=0+1-\frac{2\mathrm{x}}{60} \\ \Rightarrow \mathrm{h}\text{'}\left(\mathrm{x}\right)=1-\frac{\mathrm{x}}{30} \\ \mathrm{For} \mathrm{maixma} \mathrm{or} \mathrm{minim}, \mathrm{h}\text{'}\left(\mathrm{x}\right)=0 \\ 1-\frac{\mathrm{x}}{30}=0 \\ \Rightarrow \frac{\mathrm{x}}{30}=1 \\ \Rightarrow \mathrm{x}=30 \mathrm{metres} \\ \mathrm{h}\text{'}\left(\mathrm{x}\right)=1-\frac{\mathrm{x}}{30} \\ \mathrm{Again} \mathrm{differentiating} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x}, \mathrm{we} \mathrm{get}: \\ \mathrm{h}"\left(\mathrm{x}\right)=-\frac{1}{30} \mathrm{which} \mathrm{is} \mathrm{negative}, \mathrm{hence} \mathrm{h}\left(\mathrm{x}\right) \mathrm{will} \mathrm{be} \mathrm{maximum} \mathrm{at} \mathrm{x}=30 \\ \mathrm{h}\left(\mathrm{x}\right)=60+\mathrm{x}-\frac{{\mathrm{x}}^2}{60} \\ \Rightarrow \mathrm{h}\left(30\right)=60+30-\frac{900}{60}=90-15=75 \mathrm{m} \\ \mathrm{Height} \mathrm{of} \mathrm{building} \mathrm{is} 60 \mathrm{m}, \mathrm{hence} \mathrm{maximum} \mathrm{height} \mathrm{reached} \mathrm{by} \mathrm{ball} \mathrm{from} \\ \mathrm{ground} \mathrm{is} \mathrm{given} \mathrm{by}: \\ {\mathrm{h}}_{\max }=60+75=135 \mathrm{m} \end{gathered}$$
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