Answer: 8.73 ft/s
Step-by-step explanation:
We have the following equation that models the motion of the rock:
[tex]h(t)=-16t^{2}+64t+180[/tex]
Where [tex]h(t)[/tex] is the height of the rock at a time [tex]t[/tex].
Now, if [tex]h(t)=48 ft[/tex] we can find [tex]t[/tex] and then the velocity of the rock at that height:
[tex]48=-16t^{2}+64t+180[/tex]
Rearranging the equation:
[tex]-16t^{2}+64t+132=0[/tex]
Multiplying the equation by -1:
[tex]16t^{2}-64t-132=0[/tex]
Solving with the quadratic formula [tex]t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex] , where [tex]a=16[/tex], [tex]b=-64[/tex], [tex]c=-132[/tex] .
[tex]t=\frac{-(-64)\pm\sqrt{(-64)^{2}-4(16)(-132)}}{2(16)}[/tex]
Choosing the positive result of the equation:
[tex]t=5.5 s[/tex]
Since velocity [tex]V[/tex] is defined as the traveled distance [tex]h(t)[/tex] in a given time [tex]t[/tex], we have:
[tex]V=\frac{h(t)}{t}[/tex]
[tex]V=\frac{48 ft}{5.5 s}[/tex]
[tex]V=8.727 ft/s \approx 8.73 ft/s[/tex] This is the velocity of the rock at the height of 48 feet
