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oksian1 [2.3K]
3 years ago
12

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.
Mathematics
1 answer:
ratelena [41]3 years ago
6 0

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.

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