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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

2 Answers

5 votes

Answer: c) 64 feet

Explanation:

User Muhammad Younas
by
7.9k points
5 votes

Answer:

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)=\frac{-32}{2(-16)


(-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.

User Samarendra
by
8.6k points
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