A toy rocket is shot vertically into the air from a launching pad 4 feet above the ground with an initial velocity of 168 feet per second. The height h, in feet,
of the rocket above the ground at t seconds after launch is given by the function h(t)= -16t^2 +168t + 4. How long will it take the rocket to reach its
maximum height? What is the maximum height?

The height of the rocket as a function of time is given by the quadratic equation $ h(t) = -16t^2 + 168t + 4 $. To find the time it takes to reach the maximum height, we need to find the vertex of the parabola represented by this equation. The vertex form of a parabola is $ h(t) = a(t-h)^2 + k $, where $ (h, k) $ is the vertex of the parabola.

The time $ t $ at which the maximum height is reached is given by the formula $ t = -\frac{b}{2a} $, where $ a $ and $ b $ are the coefficients from the standard form of the quadratic equation $ h(t) = at^2 + bt + c $.

For the given function $ h(t) = -16t^2 + 168t + 4 $, $ a = -16 $ and $ b = 168 $. Plugging these values into the formula gives us:

$ t = -\frac{168}{2(-16)} = \frac{168}{32} = 5.25 \text{ seconds} $

So, it will take the rocket 5.25 seconds to reach its maximum height.

To find the maximum height, we substitute $ t = 5.25 $ back into the original height function:

$ h(5.25) = -16(5.25)^2 + 168(5.25) + 4 $

Calculating this gives us:

$ h(5.25) = -16(27.5625) + 168(5.25) + 4 $

$ h(5.25) = -441 + 882 + 4 $

$ h(5.25) = 445 \text{ feet} $

Therefore, the maximum height the rocket will reach is 445 feet.

msm555 msm555 · Answer 2 · 2024-03-15T22:37:59-04:00

Answer:

It takes 5.25 seconds to reach maximum height.

Maximum height: 445 feet

Step-by-step explanation:

To find the time it takes for the rocket to reach its maximum height, we need to find the vertex of the quadratic function representing the height of the rocket. The vertex of a quadratic function [tex]\sf f(t) = at^2 + bt + c [/tex] is given by the formula:

[tex]\Large\boxed{\boxed{\sf t = -\dfrac{b}{2a}}} [/tex]

In this case, our function is [tex]\sf h(t) = -16t^2 + 168t + 4 [/tex], so [tex]\sf a = -16 [/tex] and [tex]\sf b = 168 [/tex].

[tex]\sf t = -\dfrac{168}{2 \times (-16)} = -\dfrac{168}{-32} = 5.25 [/tex]

So, it will take the rocket [tex]\sf t = 5.25 [/tex] seconds to reach its maximum height.

To find the maximum height, we substitute [tex]\sf t = 5.25 [/tex] into the function [tex]\sf h(t) [/tex]:

[tex]\sf h(5.25) = -16(5.25)^2 + 168(5.25) + 4 [/tex]

[tex]\sf h(5.25) = -16(27.5625) + 882 + 4 [/tex]

[tex]\sf h(5.25) = -441 + 882 + 4 [/tex]

[tex]\sf h(5.25) = 445 [/tex]

So, the maximum height of the rocket is [tex]\sf 445 [/tex] feet.