a stone is thrown straight up from the edge of a roof, 600 feet above the ground, at a speed of 12 feet per second.
a. remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 4 seconds later?
b. at what time does the stone hit the ground?
c. what is the velocity of the stone when it hits the ground?

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Accepted Answer

{ "quality": "clear", "question_count": 3, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Antiderivatives", "Position Velocity Acceleration Relations" ], "new_concepts": [], "current_concepts": [ "Antiderivatives", "Position Velocity Acceleration Relations" ] } ## Formulate equations for acceleration, velocity, and position $$ \begin{aligned} a(t) &= -32 \ v(t) &= \int a(t) \, dt = -32t + C_1 \ v(0) &= 12 \implies v(t) = -32t + 12 \ s(t) &= \int v(t) \, dt = -16t^2 + 12t + C_2 \ s(0) &= 600 \implies s(t) = -16t^2 + 12t + 600 \end{aligned} $$ ## Calculate height at 4 seconds (Part A) $$ \begin{aligned} s(4) &= -16(4)^2 + 12(4) + 600 \ s(4) &= -16(16) + 48 + 600 \ s(4) &= -256 + 48 + 600 = 392 \end{aligned} $$ ## Determine impact time and velocity (Parts B and C) $$ \begin{aligned} s(t) &= 0 \implies -16t^2 + 12t + 600 = 0 \ &-4(4t^2 - 3t - 150) = 0 \ t &= \frac{3 \pm \sqrt{(-3)^2 - 4(4)(-150)}}{2(4)} = \frac{3 \pm \sqrt{9 + 2400}}{8} = \frac{3 + \sqrt{2409}}{8} \approx 6.5102 \ v(6.5102) &= -32(6.5102) + 12 \approx -196.326 \end{aligned} $$ ## Question A 392 ## Question B $\frac{3 + \sqrt{2409}}{8}$ ## Question C $12 - 4\sqrt{2409}$ { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Antiderivatives" ] }

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Question

Explanation:

Formulate equations for acceleration, velocity, and position

Calculate height at 4 seconds (Part A)

Determine impact time and velocity (Parts B and C)

Answer:

Question A

Question B

Question C