A rocket is launched in the air and can be modelled by the following equation, where h is the height in feet and the t is the time in seconds from the launch: h(t) = -10t? + 25t.

Question

a) When will the rocket return to the ground?

b) What is the maximum height of the rocket?

c) After 1 second, how high will the flare be in the air?

Answer 1

h(t) = -10t^2 +25t

it hits the ground when h(t)=0

-10t^2 +25t = 0

2t^2 -5t = 0

factor

t(2t-5) = 0

set each factor = 0, solve for t

t =0 or 5/2= 2.5 seconds to hit ground

max height is when the derivative = 0

h(t)=-10t^2+25t

h'(t) = -20t +25= 0

t = 25/20 = 1.25 seconds which is exactly half the time to hit the ground

max height = h(5/4) = -10(5/4)^2 +25(5/4)

= -10(5/4)^2 + 25(5/4)

= -10(25/16)+125/4

=-5(25/8) + 125/4

= -125/8+125/4

=-125/8+250/8

= 125/8 = 15 5/8 feet

in one second the rocket reaches

h(1) =-10(1)^2 + 25(1) = 25-10 = 15 feet

h(t)= -10t^2 +25t

rewrite in vertex form of a parabola

-10(t^2 -25t/10 + 1.25^2) + 10(25/16)

-10(t-5/2)^2 + 125/8

vertex = (5/2,125/8) in 2 1/2 seconds, max height = 15 5/8