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Hitman42 [59]
3 years ago
5

An object is launched from a launching pad 208 ft. above the ground at a velocity of 192ft/sec. what is the maximum height reach

ed by the rocket?
Mathematics
1 answer:
loris [4]3 years ago
8 0

Answer:

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

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To find the length of this segment with endpoints Y(2, 8) and Z(-2, 5), let's use the Distance Formula:

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<h2>Learn more:</h2>

Distance Formula: brainly.com/question/10134840

#LearnWithBrainly

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