URGENT SERIOUS HELP NEEDED

Question

A model rocket is shot out of a building and into the sky. The path of the rocket can be modelled by the equation ℎ = −2(𝑡 − 4)^2 + 50, where h is the height above ground in metres and t is the time in seconds since the rocket was launched. 

Determine when the height of the rocket is 32m. 

Answer 1

The equation h = -2 (t - 4)² + 50 is a parabola (because it's a squared power, and because that's what we always see for rockets and projectiles and things thrown in the air).

I think there are a couple of useful values of t we should try first.

t=0 should be useful: what height was rocket launched from?

h = -2(-4)² + 50 = -32 + 50 = 18 meters.

When is the squared term zero? When t = 4. So t = 4 also good: what is maximum height of rocket?

h = -2(0)² + 50 = 50 meters

And finally, when is the height at 32 m? Well, we just figured out it's between 0 and 4, and just under half way... or around 2 seconds as a guesstimate.

But... it's a parabola... so that's on the way up! We'll also have this height on the way down... so maybe around 6 seconds as a second guesstimate.

Let's find actual values by solving equation when h = 32m.

32 = -2(t - 4)² + 50

=> -18 = -2(t - 4)²

=> 9 = (t - 4)²

=> 3 = t - 4 and -3 = t - 4 (because of negative/positive square roots)

=> t = 7 and t = 1

Hence our rocket is at 32m of height at 1 second and again at 7 seconds.

Answer 2

At what time is h = 32. Substitute 32 for h and solve for t.

-2(t - 4)² +50 =32

-2(t-4)² = -18 (subtract 50)

(t-4)² = 9 (divide by -2)

t-4 = + 3 or t - 4 = -3 (square root property)

t = 7 or t = 1 (add 4).

It appears there are two points in time when the rocket is 32 feet above ground. Check out the following graph to confirm.

desmos.com/calculator/8jpz5yubcg