Rocker Rocket!

Algebra Level 2

A model rocket is launched from the roof of a building. It's flight path is modelled by h = − 5 t 2 + 30 t + 10 h = -5t^2+30t+10 , where h h is the height of the rocket above the ground in metres in t t seconds.

What is the rocket's maximum height?


The answer is 55.

Anish Harsha by Anish Harsha , 20, India

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2 solutions

h = − 5 t 2 + 30 t + 10 h=-5t^2+30t+10

Differentiate both sides with respect to t t .

d h d t = − 10 t + 30 \dfrac{dh}{dt}=-10t+30

For h h to be maximum, d h d t \dfrac{dh}{dt} must be equal to 0 0 . So

10 t = 30 10t=30

t = 3 t=3

Now, substitute

h = − 5 ( 3 2 ) + 30 ( 3 ) + 10 = h=-5(3^2)+30(3)+10= 55 \boxed{55}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Hung Woei Neoh
Jun 5, 2016

The rocket's height follows a quadratic function. Maximum/minimum values of quadratic functions f ( x ) = a x 2 + b x + c f(x) = ax^2+bx+c can be showed with 3 3 methods:

  1. The maximum/minimum point of a quadratic function is defined as ( − b 2 a , f ( − b 2 a ) ) \left(-\dfrac{b}{2a},f\left(-\dfrac{b}{2a}\right)\right)
  2. Complete the square into the form f ( x ) = a ( x − p ) 2 + q f(x) = a(x-p)^2 + q . The maximum/minimum point is ( p , q ) (p,q)
  3. Use Calculus. At maximum/minimum points, f ′ ( x ) = 0 f'(x) = 0

For this question, I will show Method 1

h = − 5 t 2 + 30 t + 10 h = -5t^2 + 30t+ 10

At maximum height, t = − b 2 a = − 30 2 ( − 5 ) = 3 t=-\dfrac{b}{2a} = -\dfrac{30}{2(-5)} = 3

When t = 3 , h = − 5 ( 3 2 ) + 30 ( 3 ) + 10 = − 45 + 90 + 10 = 55 t=3,\;h=-5(3^2) + 30(3) + 10 = -45+90+10 = \boxed{55}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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