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Question

A toy rocket is launched upwards such that its path can be modelled by the function h = -5t2 + 40t + 3, 

     where h is the height above the ground in metres and t is the time in seconds.

a)   What is the max height of the rocket?

b)   How long is the rocket at a height of 55m or above?

Answer 1

h(t) = -5t² + 40t + 3, where h is the height above the ground in meters and t is the time in seconds.

a.) The maximum height of the rocket is the vertex. To find the vertex, we need the axis of symmetry and plug that value in to the quadratic equation.

t = -b/2a = -40 / 2(-5) = -40 / -10 = 4

h(4) = -5(4)² + 40(4) + 3 = -5(16) + 160 + 3 = -80 + 160 + 3 = 83

The rocket will reach 83 meters above ground after 4 seconds.

b.) The rocket will reach 55 meters above ground at a certain time. We need to find any 't' that will give us 55 meters.

-5t² + 40t + 3 ≥ 55

-5t² + 40t - 52 ≥ 0

Use the quadratic formula to solve for t. Let a = -5, b = 40, and c = -52.

t = [-b ± √(b² - 4ac)] / 2a = [-40 ± √(40)² - 4(-5)(-52)] / 2(-5) = [-40 ± √(1600 - 1040)] / (-10)

t = [-40 ± √(560)] / (-10) = {1.36, 6.37}

The rocket will reach the height at least 55 meters between 1.36 and 6.37 seconds.