a. What is your equation?
This is a problem of projectile motion. A projectile is an object you throw with an initial velocity and whose trajectory is determined by the effect of gravitational acceleration. The general equation in this case is described as:
[tex]h(t)=-\frac{1}{2}gt^2+v_{0}t+h_{0}[/tex]
Where:
[tex]h(t): \ height \ at \ any \ time \\ \\ g: \ acceleration \ due \ to \ gravity \ 9.8m/s^2 \ or \ 32.16ft/s^2 \\ \\ v_{0}= \ Initial \ velocity[/tex]
So:
[tex]v_{0}=64ft/s \\ \\ h_{0}=3ft[/tex]
Finally, the equation is:
[tex]h(t)=-\frac{1}{2}(32.16)t^2+(64)t+3 \\ \\ \boxed{h(t)=-16.08t^2+64t+3}[/tex]
b. How long will it take the rocket to reach its maximum height?
The rocket will reach the maximum height at the vertex of the parabola described by the equation [tex]h(t)=-16.08t^2+64t+3[/tex]. Therefore, our goal is to find [tex]t[/tex] at this point. In math, a parabola is described by the quadratic function:
[tex]f(x)=ax^2+bx+c[/tex]
So the x-coordinate of the vertex can be calculated as:
[tex]x=-\frac{b}{2a}[/tex]
From our equation:
[tex]a=-16.08 \\ \\ b=64 \\ \\ c=3[/tex]
So:
[tex]t=-\frac{64}{2(-16.08)} \\ \\ \boxed{t=1.99s}[/tex]
So the rocket will take its maximum value after 1.99 seconds.
c. What is the maximum height the rocket will reach?
From the previous solution, we know that after 1.99 seconds, the rocket will reach its maximum, so it is obvious that the maximum height is given by [tex]h(1.99)[/tex]. Thus, we can find this as follows:
[tex]H_{max}=h(1.99)=-16.08(1.99)^2+64(1.99)+3 \\ \\ \boxed{H_{max}=66.68ft}[/tex]
So the maximum height the rocket will reach is 66.68ft
d. How long is the rocket in the air?
The rocket is in the air until it hits the ground. This can be found setting [tex]h(t)=0[/tex], so:
[tex]0=-16.08t^2+64t+3 \\ \\ Applying \ quadratic \ formula: \\ \\ t_{12}=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ a=-16.08 \\ \\ b=64 \\ \\ c=3 \\ \\ t_{12}=\frac{-64 \pm \sqrt{64^2-4(-16.08)(3)}}{2(-16.08)} \\ \\ t_{1}=4.0264 \\ \\ t_{2}=-0.046[/tex]
We can't have negative value of time, so the only correct option is [tex]t_{1}=4.0264[/tex] and rounding to the nearest hundredth we have definitively:
[tex]\boxed{t=4.03s}[/tex]
