Question:
A small rocket is launched vertically upward from the edge of a cliff 80 ft off the ground at a speed of 69 ft/s. Its height (in feet) above the ground is given by [tex]h(t) = - 25t^2 + 200t + 400[/tex] where t represents time measured in seconds.
Assuming the rocket is launched at t=0, what is an appropriate domain for h?
Answer:
[tex]0 \le t \le 9.66[/tex]
Explanation:
Given
[tex]h(t) = - 25t^2 + 200t + 400[/tex]
Required
Determine the domain of h
The initial value of t is 0.
i.e.
[tex]t = 0[/tex] --- This is given in the question
To determine the final value, we simply solve for t in [tex]h(t) = - 25t^2 + 200t + 400[/tex]
Rewrite the expression:
[tex]- 25t^2 + 200t + 400=0[/tex]
Divide through by -25
[tex]\frac{-25t^2}{-25} + \frac{200t}{-25} + \frac{400}{25}=\frac{0}{-25}[/tex]
[tex]t^2 - 8t - 16 = 0[/tex]
Using quadratic formula:
[tex]t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]
Where
[tex]a= 1[/tex] [tex]b = -8[/tex] and [tex]c = -16[/tex]
The expression becomes:
[tex]t = \frac{-(-8)\±\sqrt{(-8)^2 - 4*1*(-16)}}{2*1}[/tex]
[tex]t = \frac{8\±\sqrt{64+ 64}}{2}[/tex]
[tex]t = \frac{8\±\sqrt{128}}{2}[/tex]
[tex]\sqrt{128} = 11.31[/tex]
So, we have:
[tex]t = \frac{8\±11.31}{2}[/tex]
Split
[tex]t = \frac{8+11.31}{2}\ or\ t = \frac{8-11.31}{2}[/tex]
[tex]t = \frac{19.31}{2}\ or\ t = \frac{-3.31}{2}[/tex]
[tex]t = 9.66\ or\ t = -1.66[/tex]
But time can't be negative.
So:
[tex]t = 9.66[/tex]
Hence, the domain is:
[tex]0 \le t \le 9.66[/tex]
