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The path of a model rocket can be represented by the equation h(t) = - t ^ 2 + 20t + 12 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?

User Eridania
by
8.0k points

1 Answer

2 votes

Answer:

The height of the rocket after 3 seconds is 63 feet.

Explanation:

The path of a model rocket is represented by:


h(t)=-t^2+20t+12

where
h(t) represents the height in feet of rocket and
t represents time in seconds.

To find the height of the model rocket after 3 seconds from launch.

Solution:

In order to find the height of the model rocket after 3 seconds from launch, we will plugin
t=3 in the given function of path of rocket.

Thus, we have


h(3)=-(3)^2+20(3)+12


h(3)=-9+60+12


h(3)=63

Thus, height of the rocket after 3 seconds is 63 feet.

User WalterF
by
7.5k points
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