Question

An object is launched straight into the air. The projectile motion of the object can be modeled using h(t) = 96t – 16t2, where t is the time since launch and h(t) is the height in feet of the projectile after time t in seconds.

Answer 1

The first one is 3 seconds. The second one is 6 seconds

Explanation:

Answer 2

Answer with Step-by-step explanation:

We are given that height of projectile after t seconds is given by

`h(t)=96t-16t^2`

a.h(t)=144 ft

`144=96t-16t^2`

`16t^2-96t+144=0`

`t^2-6t+9=0`

`t^2-3t-3t+9=0`

`t(t-3)-3(t-3)=0`

`(t-3)(t-3)=0`

t-3=0

t=3

After 3 s, the height of the project will be 144 feet in the air.

b.h(t)=0

`96t-16t^2=0`

`16t(6-t)=0`

`16t=0\implies t=0`

`6-t=0\implies t=6`

At t=0, the initial position of projectile

At t=6 s , the projectile will hit the ground.