Question

1. an object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. the equation for the objects height s at time t seconds after launch is \\(s(t) = -4.9t^2 + 19.6t + 58.8\\), where s is in meters.

what is the height above the ground when the object is launched?
how long before the object hits the ground?
what is the maximum height of the object?
when does the object reach is maximum height?

1. an object is launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. the equation that models this is given by \\(h(t) = -16t^2 + 64t + 80\\).

what will be the objects maximum height?
when will it attain this height?

1. an object is launched from ground level directly upward at 39.2 m/s. the equation that models this is given by \\(s(t) = -4.9t^2 + 39.2t\\)

for how long is the object at or above a height of 34.3 meters?

Explanation:

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<reasoning>

Analyze Question 5

We are given the height equation:
\[s(t) = -4.9t^2 + 19.6t + 58.8\]

• Initial height: Find \(s(0)\).

\[s(0) = 58.8\text{ m}\]

• Time to hit the ground: Solve \(s(t) = 0\).

\[-4.9t^2 + 19.6t + 58.8 = 0\]
Divide the entire equation by \(-4.9\):
\[t^2 - 4t - 12 = 0\]
Factor the quadratic equation:
\[(t - 6)(t + 2) = 0\]
Since time must be positive, \(t = 6\text{ s}\).

• Time to reach maximum height:

Using the Vertex of Quadratic formula, \(t = -\frac{b}{2a}\):
\[t = -\frac{19.6}{2(-4.9)} = \frac{19.6}{9.8} = 2\text{ s}\]

• Maximum height: Find \(s(2)\).

\[s(2) = -4.9(2)^2 + 19.6(2) + 58.8\]
\[s(2) = -19.6 + 39.2 + 58.8 = 78.4\text{ m}\]

Analyze Question 6

We are given the height equation:
\[h(t) = -16t^2 + 64t + 80\]

• Time to reach maximum height:

Using the Vertex of Quadratic formula, \(t = -\frac{b}{2a}\):
\[t = -\frac{64}{2(-16)} = 2\text{ s}\]

• Maximum height: Find \(h(2)\).

\[h(2) = -16(2)^2 + 64(2) + 80\]
\[h(2) = -64 + 128 + 80 = 144\text{ ft}\]

Analyze Question 7

We are given the height equation:
\[s(t) = -4.9t^2 + 39.2t\]

We want to find the time interval where the height is at or above \(34.3\text{ m}\):
\[-4.9t^2 + 39.2t \ge 34.3\]
Rearrange into standard form:
\[-4.9t^2 + 39.2t - 34.3 \ge 0\]
Divide by \(-4.9\) (and reverse the inequality sign):
\[t^2 - 8t + 7 \le 0\]
Factor the quadratic expression:
\[(t - 1)(t - 7) \le 0\]
The object is at or above \(34.3\text{ m}\) between \(t = 1\text{ s}\) and \(t = 7\text{ s}\).
The duration of this interval is:
\[7 - 1 = 6\text{ s}\]
</reasoning>

<answer>

Question 5

• What is the height above the ground when the object is launched?

\(58.8\text{ meters}\)

• How long before the object hits the ground?

\(6\text{ seconds}\)

• What is the maximum height of the object?

\(78.4\text{ meters}\)

• When does the object reach its maximum height?

\(2\text{ seconds}\)

Question 6

• What will be the object's maximum height?

\(144\text{ feet}\)

• When will it attain this height?

\(2\text{ seconds}\)

Question 7

• For how long is the object at or above a height of 34.3 meters?

\(6\text{ seconds}\)
</answer>

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Answer:

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<reasoning>

Analyze Question 5

We are given the height equation:
\[s(t) = -4.9t^2 + 19.6t + 58.8\]

• Initial height: Find \(s(0)\).

\[s(0) = 58.8\text{ m}\]

• Time to hit the ground: Solve \(s(t) = 0\).

\[-4.9t^2 + 19.6t + 58.8 = 0\]
Divide the entire equation by \(-4.9\):
\[t^2 - 4t - 12 = 0\]
Factor the quadratic equation:
\[(t - 6)(t + 2) = 0\]
Since time must be positive, \(t = 6\text{ s}\).

• Time to reach maximum height:

Using the Vertex of Quadratic formula, \(t = -\frac{b}{2a}\):
\[t = -\frac{19.6}{2(-4.9)} = \frac{19.6}{9.8} = 2\text{ s}\]

• Maximum height: Find \(s(2)\).

\[s(2) = -4.9(2)^2 + 19.6(2) + 58.8\]
\[s(2) = -19.6 + 39.2 + 58.8 = 78.4\text{ m}\]

Analyze Question 6

We are given the height equation:
\[h(t) = -16t^2 + 64t + 80\]

• Time to reach maximum height:

Using the Vertex of Quadratic formula, \(t = -\frac{b}{2a}\):
\[t = -\frac{64}{2(-16)} = 2\text{ s}\]

• Maximum height: Find \(h(2)\).

\[h(2) = -16(2)^2 + 64(2) + 80\]
\[h(2) = -64 + 128 + 80 = 144\text{ ft}\]

Analyze Question 7

We are given the height equation:
\[s(t) = -4.9t^2 + 39.2t\]

We want to find the time interval where the height is at or above \(34.3\text{ m}\):
\[-4.9t^2 + 39.2t \ge 34.3\]
Rearrange into standard form:
\[-4.9t^2 + 39.2t - 34.3 \ge 0\]
Divide by \(-4.9\) (and reverse the inequality sign):
\[t^2 - 8t + 7 \le 0\]
Factor the quadratic expression:
\[(t - 1)(t - 7) \le 0\]
The object is at or above \(34.3\text{ m}\) between \(t = 1\text{ s}\) and \(t = 7\text{ s}\).
The duration of this interval is:
\[7 - 1 = 6\text{ s}\]
</reasoning>

<answer>

Question 5

• What is the height above the ground when the object is launched?

\(58.8\text{ meters}\)

• How long before the object hits the ground?

\(6\text{ seconds}\)

• What is the maximum height of the object?

\(78.4\text{ meters}\)

• When does the object reach its maximum height?

\(2\text{ seconds}\)

Question 6

• What will be the object's maximum height?

\(144\text{ feet}\)

• When will it attain this height?

\(2\text{ seconds}\)

Question 7

• For how long is the object at or above a height of 34.3 meters?

\(6\text{ seconds}\)
</answer>

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"question_type": "Multi-part",
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