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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.

2 Answers

3 votes

Answer:

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

Explanation:

User MOHAMED
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1 vote

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.

User Mordaroso
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