An improvised rocket is launched from the top of a building $150 \text{ feet}$ high. The height $\text{ (H) }$ reached by the rocket after $\text{ (t) }$ seconds is given by the function:

$\large H( t ) =40t-10{ t }^{ 2 }$

How long (in seconds) did it take to reach the highest point?

The answer is 2.

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$\begin{aligned} H(t) & = 40t - 10t^2\\ H'(t) & = \dfrac{d}{dt}(40t - 10t^2)\\ H & = 40 - 20t\\ \text{Now for maximum height, equate to 0}\\ 40 - 20t & = 0\\ t & = \boxed{2}sec \end{aligned}$