Questions: The height s (in feet) at time t (in seconds) of a silver dollar dropped from the top of a building is given by s=-16 t^2+585. (a) Find the average velocity on the interval [4,5]. ft / s (b) Find the instantaneous velocities when t=4 and when t=5. s'(4)= ft / s s'(5)= ft / s (c) How long will it take the dollar to hit the ground? (Round your answer to two decimal places.) s (d) Find the velocity of the dollar when it hits the ground. (Round your answer to one decimal place.) ft / s

Solution Steps

(a) To find the average velocity on the interval \([4,5]\), calculate the change in height over the change in time. This involves evaluating the height function at \(t=4\) and \(t=5\), and then using the formula for average velocity: \(\frac{s(5) - s(4)}{5 - 4}\).

(b) The instantaneous velocity at a given time is the derivative of the height function evaluated at that time. First, find the derivative of the height function \(s(t)\), and then evaluate it at \(t=4\) and \(t=5\).

(c) To find how long it will take the dollar to hit the ground, solve for \(t\) when the height \(s(t) = 0\).

The average velocity is calculated using the formula:

\[ \text{Average Velocity} = \frac{s(5) - s(4)}{5 - 4} \]

Given the height function \(s(t) = -16t^2 + 585\), we find:

\[ s(4) = 585 - 16 \times 4^2 = 329 \] \[ s(5) = 585 - 16 \times 5^2 = 185 \]

Thus, the average velocity is:

\[ \frac{185 - 329}{5 - 4} = -144 \, \text{ft/s} \]

The instantaneous velocity is the derivative of the height function evaluated at the given time. The derivative of \(s(t)\) is:

\[ s'(t) = -32t \]

Evaluating at \(t=4\):

\[ s'(4) = -32 \times 4 = -128 \, \text{ft/s} \]

Evaluating at \(t=5\):

\[ s'(5) = -32 \times 5 = -160 \, \text{ft/s} \]

To find when the dollar hits the ground, solve for \(t\) when \(s(t) = 0\):

\[ -16t^2 + 585 = 0 \]

Solving for \(t\), we get:

\[ t = \pm \frac{3\sqrt{65}}{4} \]

Since time cannot be negative, we take the positive value:

\[ t = \frac{3\sqrt{65}}{4} \approx 6.0311 \, \text{seconds} \]

Final Answer