Answer:
a) 64 feet
b) 3 seconds
Step-by-step explanation:
a)
The maximum height of [tex]h=h(t)[/tex] can be bound by finding the y-coordinate of the vertex of [tex]y=-16x^2+32x+48[/tex].
Compare this equation to [tex]y=ax^2+bx+c[/tex] to find the values of [tex]a,b,\text{ and } c[/tex].
[tex]a=-16[/tex]
[tex]b=32[/tex]
[tex]c=48[/tex].
The x-coordinate of the vertex can be found by evaluating:
[tex]\frac{-b}{2a}=\frac{-32}{2(-16)[/tex]
[tex]\frac{-b}{2a}=\frac{-32}{-32}[/tex]
[tex]\frac{-b}{2a}=1[/tex]
So the x-coordinate of the vertex is 1.
The y-coordinate can be found be evaluating [tex]y=-16x^2+32x+48[/tex] at [tex]x=1[/tex]:
[tex]y=-16(1)^2+32(1)+48[/tex]
[tex]y=-16+32+48[/tex]
[tex]y=16+48[/tex]
[tex]y=64[/tex]
So the maximum height of the rocket is 64 ft high.
b)
When the rocket hit's the ground the height that the rocket will be from the ground is 0 ft.
So we are trying to find the second t such that:
[tex]0=-16t^2+32t+48[/tex]
I'm going to divide both sides by -16:
[tex]0=t^2-2t-3[/tex]
Now we need to find two numbers that multiply to be -3 and add to be -2.
Those numbers are -3 and 1 since (-3)(1)=-3 and (-3)+(1)=-2.
[tex]0=(t-3)(t+1)[/tex]
This implies we have either [tex]t-3=0[/tex] or [tex]t+1=0[/tex]
The first equation can be solved by adding 3 on both sides: [tex]t=3[/tex].
The second equation can be solved by subtracting 1 on both sides: [tex]t=-1[/tex].
So when [tex]t=3[/tex] seconds, is when the rocket has hit the ground.
