Question

Mose decides to launch a rocket out of the barn at Schrute farms. The equation below represents the height (in meters), of the rocket getting launched above the ground, t seconds after the rocket is launched. h(t) = -.4 (t - 2)^2 + 5 How high above the ground is the rocket before it is launched?

Answer 1

Height at an instance of time is given by :

h(t) = -4( t - 2 )² + 5     .....1

Differentiating above equation with respect to t .

$\dfrac{dh}{dt}= -4\times ( -2 ) ( t - 2 ) \\\\\dfrac{dh}{dt}= 8(t-2)$

Now, for maximum , $\dfrac{dh}{dt} = 0$ .

8( t - 2 ) = 0

t = 2

Putting value of t = 2 in given equation, we get :

h( 2 ) = -4( 2 - 2 )² + 5

h( 2 ) = 5 m

Therefore, maximum height rocket can reach is 5 m.