Question

A person standing at the edge of a cliff throws a rock upward from a height of 180ft above ground level with an initial velocity of 64ft/s. Calculate the velocity of the rock when it is exactly 48ft above the person’s hand. Separate multiple answers with a comma. (Hint: Use h(t)=−16t2+64t+180 as the position function, where h is in feet, t in seconds. Ignore air resistance.)

Answer 1

Answer: 8.73 ft/s

Step-by-step explanation:

We have the following equation that models the motion of the rock:

$h(t)=-16t^{2}+64t+180$

Where $h(t)$ is the height of the rock at a time $t$.

Now, if  $h(t)=48 ft$ we can find $t$ and then the velocity of the rock at that height:

$48=-16t^{2}+64t+180$

Rearranging the equation:

$-16t^{2}+64t+132=0$

Multiplying the equation by -1:

$16t^{2}-64t-132=0$

Solving with the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$  , where $a=16$, $b=-64$, $c=-132$ .

$t=\frac{-(-64)\pm\sqrt{(-64)^{2}-4(16)(-132)}}{2(16)}$

Choosing the positive result of the equation:

$t=5.5 s$  

Since velocity $V$ is defined as the traveled distance $h(t)$ in a given time $t$, we have:

$V=\frac{h(t)}{t}$

$V=\frac{48 ft}{5.5 s}$

$V=8.727 ft/s \approx 8.73 ft/s$ This is the velocity of the rock at the height of 48 feet