Question

A rocket is launched into the air. The projectile motion of the rocket can be modeled using h(t) = 112t –16t2, where t is the time since launch in seconds and h(t) is the height of the rocket at time t. When will the rocket be 196 feet in the air?

Answer 1

After 3.5 sec rocket will be at 196 feet in the air.

Explanation:

The projectile motion of a rocket can be modeled using h(t) = 112t - 16t²

where t is the time since launch in seconds.

h(t) is the height of the rocket at time t.

we have to calculate when height h is 196 feet.

Now from the given function

196 = 112t - 16t²

-16t² + 112t - 196 = 0

-4t² + 28t - 49 = 0

Or 4t² - 28t + 49 = 0

(2t - 7)² = 0

2t - 7 = 0

t = 7/2 = 3.5 sec

Answer 2

h ( t ) = 112 t - 16 t²
When the rocket is 196 feet in the air:
196 = 112 t - 16 t²
16 t² - 112 t + 196 = 0 / : 4
4 t² - 28 t + 49 = 0
( 2 t - 7 )² = 0
2 t - 7 = 0
2 t = 7
t = 7 : 2
t = 3.5 s
Answer: The rocket will be 196 feet in the air after 3.5 seconds.