110th Problem 2016

Algebra Level 1

An improvised rocket is launched from the top of a building 150 feet 150 \text{ feet} high. The height (H) \text{ (H) } 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 }

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

Check out the set: 2016 Problems .


The answer is 2.

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 ′ ( 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 \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}

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I think your notation is wrong. Shouldn't it be d d t ( H ( t ) ) \dfrac{d}{dt}\left(H(t)\right) or H ′ ( t ) H'(t) ?

Hung Woei Neoh - 5 years, 1 month ago

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Oh yes, I didnt write H inside thanks! :)

Ashish Menon - 5 years, 1 month ago

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