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Mathematics
28.09.2021 22:57
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An object is launched from a launching pad 208 ft. above the ground at a velocity

An object is launched from a launching pad 208 ft. above the ground at a velocity of 192ft/sec. what is the maximum height reached by the rocket?
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sarn8899
sarn8899
4,5(74 marks)

The maximum height is 784 feet

Step-by-step explanation:

In this problem we use the kinematic equation of the height h of an object as a function of time

h(t) = -16t ^ 2 + v_0t + h_0

Where v_0 is the initial velocity and h_0 is the initial height.

We know that

v_0 = 192\ \frac{ft}{sec}

h_0 = 208\ ft.

Then the equation of the height is:

h(t) = -16t ^ 2 + 192t +208

For a quadratic function of the form ax ^ 2 + bx + c

where a

the maximum height of the function is at its vertex.

The vertice is

x = -\frac{b}{2a}\\\\y = f(\frac{-b}{2a})

In this case

a = -16\\b = 192\\c = 208

Then the vertice is:

t = -\frac{192}{2(-16)}\\\\t = 6\ sec

Now we calculate h (6)

h(6) = -16(6) ^ 2 +192(6) +208\\\\h(6) = 784\ feet

The maximum height is 784 feet

annabel11
annabel11
4,7(67 marks)

\frac{ {x}^{7} + 2x - 64 }{x}

Step-by-step explanation:

{x}^{6} - 64 \div x + 2

➡️ {x}^{6} - \frac{64}{x} + 2

➡️ \frac{ {x}^{7} - 64 + 2x }{x}

➡️ \frac{ {x}^{7} + 2x - 64 }{x}

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