Question

A rock is dropped off the edge of a cliff and its distance s (in feet) from the top of the cliff after t seconds is s(t) = 16t2. Assume the distance from the top of the cliff to the water below is 96 ft.
A) When will the rock strike the water?
B) Make a table of average velocities and approximate the velocity at which the rock strikes the water.

Answer 1

Answer:

A) 2.45 seconds

B)$\begin{center}\begin{tabular}{ c c}\bold{Average Velocity} & \bold{Velocity at strike of water} \\ 39.18 & 78.4 \\ \end{tabular}\end{center}$

Step-by-step explanation:

Distance $s$ (in feet) from the top of cliff after $t$ seconds when a rock is dropped off the edge of a cliff is given as:

$s(t)=16t^2$

Distance from the top of cliff to water = 96 ft

To find:

A) When will the rock strike the water

i.e. The time taken by the rock to travel the distance of 96 ft.

B) Table of Average Velocity and approximate velocity at which the rock strikes the water ?

Solution:

A) Given that distance traveled = 96 ft

To find the value of $t$

Putting the value of $s$ in formula: $s(t)=16t^2$ to find $t$:

$96 = 16t^2\\\Rightarrow t^2=6\\\Rightarrow \bold{t = 2.45\ sec}$

B) The velocity at which the rock hits the water:

We need to differentiate $s(t)$ w.r.to $t$ in order to find the velocity at any instant.

$s'(t) =\dfrac{d}{dt}s(t) = \dfrac{d}{dt}(16t^2)\\\Rightarrow v = 2\times 16t^{(2-1)}\\\Rightarrow v = 32t$

The time taken to reach the water is 2.45 seconds.

Putting $t$ = 2.45 seconds to find the velocity:

$v = 32\times 2.45 = 78.4\ ft/sec$

Average velocity is given as total distance traveled and total time taken.

$\text{Average Velocity = }\dfrac{\text{Total Distance traveled}}{\text{Total Time Taken}}$

$\text{Average Velocity = }\dfrac{96}{2.45}\\\text{Average Velocity = 39.18\ ft/s}$

$\begin{center}\begin{tabular}{ c c}\bold{Average Velocity} & \bold{Velocity at strike of water} \\ 39.18 & 78.4 \\ \end{tabular}\end{center}$