Questions: A rock is launched vertically upward from the ground with a speed of 96 ft / s. The height of the rock (in ft ) above the ground after t seconds is given by the function s(t)=-16 t^2+96 t. Find its average velocity during its flight. The average velocity of the rock during its flight is (Simplify your answer.) ft / sec.

A rock is launched vertically upward from the ground with a speed of 96 ft / s. The height of the rock (in ft ) above the ground after t seconds is given by the function s(t)=-16 t^2+96 t. Find its average velocity during its flight.

The average velocity of the rock during its flight is (Simplify your answer.) ft / sec.

Transcript text: A rock is launched vertically upward from the ground with a speed of $96 \mathrm{ft} / \mathrm{s}$. The height of the rock (in ft ) above the ground after t seconds is given by the function $s(t)=-16 t^{2}+96 t$. Find its average velocity during its flight. The average velocity of the rock during its flight is $\square$ $\square$ (Simplify your answer.) $\mathrm{ft} / \mathrm{sec}$.

Solution

Solution Steps

Step 1: Determine the total flight time

The rock's height function is given by $ s(t) = -16t^2 + 96t $.
To find the total flight time, set $ s(t) = 0 $ and solve for $ t $: \[ -16t^2 + 96t = 0 \]
Factor out $ t $: \[ t(-16t + 96) = 0 \]
Solve for $ t $: \[ t = 0 \quad \text{or} \quad t = 6 \]
The total flight time is $ 6 $ seconds.

Step 2: Calculate the initial and final heights

The initial height $ s(0) $ is: \[ s(0) = -16(0)^2 + 96(0) = 0 \, \text{ft} \]
The final height $ s(6) $ is: \[ s(6) = -16(6)^2 + 96(6) = -576 + 576 = 0 \, \text{ft} \]

Step 3: Compute the average velocity

The average velocity $ v_{\text{avg}} $ is given by the change in height divided by the total flight time: \[ v_{\text{avg}} = \frac{s(6) - s(0)}{6 - 0} = \frac{0 - 0}{6} = 0 \, \text{ft/sec} \]

The average velocity of the rock during its flight is $ 0 \, \text{ft/sec} $.