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

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.