Answer with Step-by-step explanation:
We are given that height of projectile after t seconds is given by
[tex]h(t)=96t-16t^2[/tex]
a.h(t)=144 ft
[tex]144=96t-16t^2[/tex]
[tex]16t^2-96t+144=0[/tex]
[tex]t^2-6t+9=0[/tex]
[tex]t^2-3t-3t+9=0[/tex]
[tex]t(t-3)-3(t-3)=0[/tex]
[tex](t-3)(t-3)=0[/tex]
t-3=0
t=3
After 3 s, the height of the project will be 144 feet in the air.
b.h(t)=0
[tex]96t-16t^2=0[/tex]
[tex]16t(6-t)=0[/tex]
[tex]16t=0\implies t=0[/tex]
[tex]6-t=0\implies t=6[/tex]
At t=0, the initial position of projectile
At t=6 s , the projectile will hit the ground.
Answer:
The first one is 3 seconds. The second one is 6 seconds
Step-by-step explanation:
