When the projectile reaches
its maximum height, the slope of the projectile is horizontal at that maximum point, hence its derivative is zero.
Given
\(h(t) = - 16 {t}^{2} + 440t\)
The derivative is given by:
\(h'(t) = - 32 {t} + 440\)
At maximum height,
\(h'(t) = 0\)
\(\Rightarrow - 32 {t} + 440 =0\)
Now let us solve for t, to obtain;
\(\ - 32 {t} = - 440\)
\(\Rightarrow t = \frac{ - 440}{ - 32}\)
\(\Rightarrow t = 13.75s\)
\(Hence the projectile reached the maximum height after 13.75 seconds.\)
