Question

Ian tosses a bone up in the air for his dog, Spot. The height, h, in feet, that Spot is above the ground at the time t seconds after she jumps for the bone can be represented h(t)=-16²+20t.
What is Spot's average rate of ascent, in feet per second, from the time she jumps into the air to the time she catches the bone at t=1/2?

Answer 1

Answer: 12 ft/s

Step-by-step explanation:

We are told the following function models Spot's height at time $t$:

$h_{t}=-16t^{2}+20t$ (1)

And we are asked to find Spot's average rate of ascent, this means its velocity $V$, which is calculated by:

$V=\frac{h_{t=\frac{1}{2}}-h_{t=0}}{t_{f}-t_{o}}$ (2)

Where:

$h_{t=\frac{1}{2}$ is the height of Spot at time $t=\frac{1}{2}s$

$h_{t=0$ is the height of Spot at time $t=0 s$

$t_{f}=\frac{1}{2} s$ is the final time

$t_{o}=0 s$ is the initial time

So, firstly we need to calculate $h_{t=\frac{1}{2}$ and $h_{t=0$:

$h_{t=\frac{1}{2}}=-16(\frac{1}{2})^{2}+20(\frac{1}{2})$ (3)

$h_{t=\frac{1}{2}}=6 ft$ (4)

$h_{t=0}=0 ft$ (5)

Substituting these values in (2):

$V=\frac{6 ft-0 ft}{\frac{1}{2} s-0 s}$ (6)

Finally:

$V=12 ft/s$ This is Spot's average rate of ascent