a) 45.1 seconds
b) 2587.5 ft
Step-by-step explanation:
a)
The height of the rocket as a function of time is given by:
[tex]h(t)=-4.9t^2+217t+185[/tex]
where
t is the time (in seconds)
-4.9 represents the acceleration of gravity (in ft/s^2)
+217 represents the initial velocity (in ft/s)
+185 represents the initial heigth of the rocket at t = 0 (in feet)
Here we want to find the time t at which the rocket splashes down, so the time t at which the height of the rocket is zero:
h(t) = 0
To do so, we substituting h(t) = 0 into the expression:
[tex]0=-4.9t^2+217t+185\\4.9t^2-217t-185=0[/tex]
Using the formula to find t:
[tex]t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-217)\pm \sqrt{(-217)^2-4(4.9)(-185)}}{2(4.9)}=\frac{217\pm 225.2}{9.8}[/tex]
which gives two solutions:
[tex]t_1=45.1 s\\t_2=-0.8 s[/tex]
Neglecting the negative solution, the rocket splashes down after 45.1 seconds.
b)
In order to find the maximum height of the rocket, we have to find the time t at which the derivative of the function h(t) becomes zero.
The function is
[tex]h(t)=-4.9t^2+217t+185[/tex]
Therefore, its derivative vs t is:
[tex]h'(t)=-2\cdot 4.9 t+217=-9.8t+217[/tex]
This expression represents the velocity of the rocket at time t.
The derivative becomes zero when h'(t) = 0, so:
[tex]0=-9.8t+217\\t=\frac{217}{9.8}=22.1s[/tex]
So, the rocket reaches the maximum height after 22.1 seconds. Substituting this value into h(t), we find the maximum height reached by the rocket:
[tex]h_{max}=-4.9(22.1)^2+217(22.1)+185=2587.5 ft[/tex]
