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A toy rocket is launched from a platform that is 48 feet high. The rocket's height above the ground is modeled by the equation [tex] h = -16t^2 + 32t + 48 [/tex].

What is the rocket's height at 2 seconds?

Please show work.

Answer :

a) rocket height is 64 feet

a)

The maximum height of h = h(t) can be bound by finding the y-coordinate of the vertex of .

y = -16x²+32x+48

Compare this equation to y = ax²+bx+c to find the values of a, b, and c

a = -16

b = 32

c = 48

The x-coordinate of the vertex can be found by evaluating:

-b/2a

So the x-coordinate of the vertex is 1.

The y-coordinate can be found be evaluating at x = 1 in the equation,

We get,

y = 64

So the maximum height of the rocket is 64 ft high.

b)

When the rocket hit's the ground the height that the rocket will be from the ground is 0 ft.

So we are trying to find the second t such that:

0 = -16t²+32t+48

I'm going to divide both sides by -16:

0 = t²- 2t - 3

Now we need to find two numbers that multiply to be -3 and add to be -2.

Those numbers are -3 and 1 since (-3)(1)=-3 and (-3)+(1)=-2.

0 = (t-3)(t+1)

This implies we have either or

t - 3 = 0 or t + 1 = 0

The first equation can be solved by adding 3 on both sides: .

The second equation can be solved by subtracting 1 on both sides: t = -1

So when t = 3 seconds, is when the rocket has hit the ground.

To learn more about quadratic equation visit:

https://brainly.com/question/30098550

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