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A toy rocket is launched upward from the ground at a rate of 64 feet per second. The function [tex]h(t) = -16t^{2} + 64t[/tex] represents [tex]h[/tex], the height of the rocket, at any given time, [tex]t[/tex], in seconds after the launch.

How many seconds after launch will the rocket hit the ground?

Answer :

The rocket will hit the ground at t = 4 seconds after launch.

To find out when the rocket will hit the ground, you need to determine the time at which the height (h(t)) becomes zero. In this case, the equation for the height of the rocket is given by:

h(t) = -16t^2 + 64t

You want to find the time when h(t) = 0, so you can set the equation equal to zero and solve for t:

-16t^2 + 64t = 0

Now, you can factor out -16t to simplify the equation:

-16t(t - 4) = 0

Now, you have a product of two factors, and you can set each factor equal to zero and solve for t:

-16t = 0

Divide both sides by -16:

t = 0

t - 4 = 0

Add 4 to both sides:

t = 4

Now, you have two possible solutions: t = 0 and t = 4 seconds.

The rocket is launched from the ground at t = 0 seconds, so t = 0 represents the initial time of launch.

Therefore, the rocket will hit the ground at t = 4 seconds after launch.

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