109th Problem 2016

Algebra Level 2

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

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

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

Check out the set: 2016 Problems .


The answer is 190.

Angela Fajardo by Angela Fajardo , 19, Philippines

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1 solution

Ashish Menon
Apr 26, 2016

H ( t ) = 40 t − 10 t 2 H d t d t = d d t ( 40 t − 10 t 2 ) H = 40 − 20 t Now for maximum height, equate to 0 40 − 20 t = 0 t = 2 s e c H ( 2 ) = ( 40 × 2 ) − ( 10 × 4 ) = 80 − 40 = 40 f e e t \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}

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

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