Question

if a rock is dropped from a height of 82 ft, its position t seconds after it is dropped until it hits the ground is given by the function ( s(t)=-16 t^{2}+82 ).
round values below to 3 decimal places.
how long does it take the rock to hit the ground?
seconds
find the average velocity of the rock from when it is released until when it hits the ground.
hint: the original velocity, at time ( t = 0 ), is 0 ft/sec (because the rock is dropped, not thrown). use your answer to the first question to find the terminal velocity, then find the average velocity over the time interval.
feet per second
what time after the rock is thrown will its instantaneous velocity be equal to its average velocity? (apply the mean value theorem: set the derivative equal to the previous answer, and solve for t.)
seconds after it is thrown
question help: message instructor

Explanation:

Step1: Find the time when the rock hits the ground

When the rock hits the ground, \(s(t)=0\). So we set \(-16t^{2}+82 = 0\).

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Since \(t\geq0\), \(t=\sqrt{\frac{82}{16}}\approx2.264\) seconds.

Step2: Find the average velocity

The average velocity formula is \(v_{avg}=\frac{s(t_{2})-s(t_{1})}{t_{2}-t_{1}}\). Here, \(t_{1} = 0\), \(s(0)=-16\times0^{2}+82 = 82\), \(t_{2}\approx2.264\), \(s(2.264)=0\).
\(v_{avg}=\frac{0 - 82}{2.264-0}\approx - 36.220\) ft/s.

Another way: The velocity function \(v(t)=s^{\prime}(t)=-32t\). The terminal velocity \(v(2.264)=-32\times2.264=-72.448\) ft/s. Using the formula for average of an arithmetic - sequence (since \(v(t)\) is a linear function of \(t\) for constant - acceleration motion \(v(t)=v_{0}+at\), here \(v_{0} = 0\), \(a=-32\)), \(v_{avg}=\frac{v(0)+v(2.264)}{2}=\frac{0+( - 72.448)}{2}=-36.224\approx - 36.220\) ft/s.

Step3: Use the Mean Value Theorem

We know that \(s^{\prime}(t)=v(t)=-32t\). We want to find \(t\) such that \(v(t)=v_{avg}\).
Set \(-32t=-36.220\)

$$ t=\frac{-36.220}{-32}\approx1.132 $$

Answer:

1. \(2.264\) seconds
2. \(-36.220\) feet per second
3. \(1.132\) seconds after it is thrown