Answer:
The approximate speed of the object when it hits the ground is 63.568 meters per second.
Explanation:
The height of the object launched vertically upward is represented by the function [tex]h(t) = 200+11\cdot t -4.9\cdot t^{2}[/tex]. The approximate speed of the object when it hits the ground occurs when [tex]h(t) = 0[/tex]. Then, we solve the resulting second-order polynomial by the Quadratic Formula:
[tex]-4.9\cdot t^{2}+11\cdot t +200 = 0[/tex] (1)
The roots associated to this polynomial are, respectively:
[tex]t_{1}\approx 7.609\,s[/tex] and [tex]t_{2} \approx -5.364\,s[/tex]
Given that time is a positive variable, the only solution that is physically reasonable is:
[tex]t\approx 7.609\,s[/tex]
The function velocity is the first derivative of the function defined at the commencement of the explanation, that is:
[tex]v(t) = 11-9.8\cdot t[/tex] (2)
If we know that [tex]t\approx 7.609\,s[/tex], then the approximate velocity of the object when it hits the ground is:
[tex]v(7.609) = 11-9.8\cdot (7.609)[/tex]
[tex]v(7.609) = -63.568\,\frac{m}{s}[/tex]
The speed is the magnitude of this velocity. Therefore, the approximate speed of the object when it hits the ground is 63.568 meters per second.
