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sopheary dara mao's question at Yahoo! Answers regarding kinematics

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Feb 24, 2012
Here is the question:

Calculus Antiderivative problem.?

A stone is thrown straight up from the edge of a roof, 800 feet above the ground, at a speed of 14 feet per second.
A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 2 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?
Here is a link to the question:

Calculus Antiderivative problem.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
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Feb 24, 2012
Hello sopheary dara mao,

I would orient the vertical $x$-axis of motion such that the origin coincides with the initial position of the stone, and model the motion of the stone with the IVP (where $x$ is measured in feet and time $t$ is measured in seconds):

\(\displaystyle x''(t)=-g\) where \(\displaystyle x'(0)=v_0\) and \(\displaystyle x(0)=x_0\)

Integrating once, we find:

\(\displaystyle x'(t)=v(t)=-gt+C\)

Using the initial conditions, we find:

\(\displaystyle v_0=C\) hence:

\(\displaystyle v(t)=-gt+v_0\)

Integrating again, we find:

\(\displaystyle x(t)=-\frac{g}{2}t^2+v_0t+C\)

Using the initial conditions, we find:

\(\displaystyle x_0=C\) hence:

\(\displaystyle x(t)=-\frac{g}{2}t^2+v_0t+x_0\)

Now, using the given data and orientation of our coordinate system, i.e:

\(\displaystyle g=32,\,v_0=14,\,x_0=0\) we have:

\(\displaystyle x(t)=-16t^2+14t\)

\(\displaystyle v(t)=-32t+14\)

Now we are ready to answer the questions:

A.) \(\displaystyle x(2)=-16(2)^2+14(2)=-36\)

This means after 2 seconds the stone is 36 feet below its initial position, or 764 ft above the ground.

B.) To find at what time the stone hits the ground, we may set:

\(\displaystyle x(t)=-800\)

\(\displaystyle -16t^2+14t=-800\)

\(\displaystyle 16t^2-14t-800=0\)

\(\displaystyle 2\left(8t^2-7t-400 \right)=0\)

Applying the quadratic formula, and taking the positive root, we find:

\(\displaystyle t=\frac{7+\sqrt{12849}}{16}\approx7.5221\)

C.) \(\displaystyle v\left(\frac{7+\sqrt{12849}}{16} \right)=-32\left(\frac{7+\sqrt{12849}}{16} \right)+14=-2\sqrt{12849}\)

The negative sign indicates the stone is moving in a downward direction when it hits the ground. This is in agreement with the speed when using energy considerations, i.e., equating initial energy (gravitational potential and kinetic) to final energy (kinetic) and solving for the final velocity.

To sopheary dara mao and any other guests viewing this topic, I invite and encourage you to post other kinematics questions in either our Calculus forum if the calculus is to be used or our Other Topics forum if energy considerations, or other pre-calculus techniques are to be used.

Best Regards,