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

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

What is the highest distance of the rocket from the ground?

The answer is 190.

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$\begin{aligned} H(t) & = 40t - 10t^2\\ H \dfrac{dt}{dt} & = \dfrac{d}{dt}(40t - 10t^2)\\ H & = 40 - 20t\\ \text{Now for maximum height, equate to 0}\\ 40 - 20t & = 0\\ t & = 2sec\\ \\ H(2) & = (40 × 2) - (10 × 4)\\ & = 80 - 40\\ & = 40 feet \end{aligned}$

$\therefore \text{Total distance from ground} = 150 + 40 = \boxed{190}$ feet.