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aivan3 [116]
3 years ago
14

The path of a model rocket can be represented by the equation h(t)=-t2+15t+16, where h(t) is the height, in feet, of the rocket

at any given time, t. what is the height of the model rocket after 3 seconds from launch?
Mathematics
2 answers:
Musya8 [376]3 years ago
4 0
The height is 52 feet.

Using t=3, we have:
-3² + 15(3) + 16 = -9 + 45 + 16 = 52
Korvikt [17]3 years ago
4 0

we are given

h(t)=-t^2+15t+16

where h(t) is height in feet

t is time in seconds

now, we can plug t=3 and find height

h(3)=-(3)^2+15(3)+16

h(3)=-9+45+16

h(3)=52

so,

the height of the model rocket after 3 seconds from launch is 52 feet.......Answer

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