111th 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) } 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 distance of the rocket from the ground (in feet) after 3 seconds?


Check out the set: 2016 Problems .


The answer is 180.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Ashish Menon
Apr 26, 2016

In this question, just substitute t = 3 in the given equation.
H ( t ) = ( 40 × 3 ) ( 10 × 3 2 ) = ( 120 90 ) = 30 H(t) = (40 × 3) - (10 × 3^2) = (120 - 90) = 30 .
So, the distance from the ground is 150 + 30 = 180 150 + 30 = \boxed{180} feet.

Ehhh... why is there a bracket there?

( 120 90 = 30 ) (120-90=30)

Hung Woei Neoh - 5 years, 1 month ago

Log in to reply

Haha, LOL Thats a funny typo.. the bracket was intended just to keep after 90, thanks! I have edited it.

Ashish Menon - 5 years, 1 month ago

Log in to reply

You're welcome

Hung Woei Neoh - 5 years, 1 month ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...