The flight of one type of rocket is described by a function h = −5(t − 3)2 + 50, where h is the height of the rocket, in metres, and t is the time, in seconds, since the rocket was fired.

Question

a. What was the max height of the rocket?

b. At what time the max height was reached?

c. What was the initial height of the rocket?

d. What was the height of the rocket after 2 seconds of the flight? At what another

time this height was reached?

e. Was the rocket in the air after 6 sec?

Answer 1

max height = 50 m

max height was reached at time t=3 seconds .

initial height = h(0) = -5(3)^2 + 50 = 50-45 = 5 m

h(2) = -5+50 = 45 m after 2 seconds

h(4) also = 45m after 4 seconds

h(6) = -5(6)^2 + 50 = 50-180= -130 m into a hole in the ground. It's no longer in the air at 6 seconds

You could rewrite the equation as

h(t) = -5t^2 + 30t +5 then you can read off initial height as 5 m and initial velocity as 30 m/s

-5 should really be -4.9, but apparently the problem maker rounded it off. It's the deceleration due to gravity