Question

A toy rocket is launched from a platform that is 48 ft high. The rocket height above the ground is modeled by h=-16t^2+32t+48. a) find the maximum height of the rocket b) find the time it will take for the rocket to reach the ground

Answer 1

Answer: c) 64 feet

Explanation:

Answer 2

a) 64 feet

b) 3 seconds

Explanation:

a)

The maximum height of
`h=h(t)` can be bound by finding the y-coordinate of the vertex of
`y=-16x^2+32x+48`.

Compare this equation to
`y=ax^2+bx+c` to find the values of
`a,b,\text{ and } c`.

`a=-16`

`b=32`

`c=48`.

The x-coordinate of the vertex can be found by evaluating:

`(-b)/(2a)=(-32)/(-32)`

`(-b)/(2a)=1`

So the x-coordinate of the vertex is 1.

The y-coordinate can be found be evaluating
`y=-16x^2+32x+48` at
`x=1`:

`y=-16(1)^2+32(1)+48`

`y=-16+32+48`

`y=16+48`

`y=64`

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:

`0=-16t^2+32t+48`

I'm going to divide both sides by -16:

`0=t^2-2t-3`

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.

`0=(t-3)(t+1)`

This implies we have either
`t-3=0` or
`t+1=0`

The first equation can be solved by adding 3 on both sides:
`t=3`.

The second equation can be solved by subtracting 1 on both sides:
`t=-1`.

So when
`t=3` seconds, is when the rocket has hit the ground.