Question

Peter and Sophia launch two model rockets from different initial heights. The height, h(t), in feet, of Peter's rocket t seconds after launch is
modeled by the equation h(t)= - 16+2 +64t + 80, where -16 represents the gravity constant in feet per second squared and 64 represents the
initial velocity in feet per second.
Sophia launches her rocket from an initial height that is half the height from where Peter's rocket is launched. The initial velocity of Sophia's
rocket is twice the initial velocity of Peter's rocket.
What is the difference, in feet, between the maximum height of Peter's rocket and the maximum height of Sophia's rocket?

Answer 1

Using quadratic equations, it is found that the difference between the maximum heights is of 152 feet.

The height of a rocket after t seconds is given by:

$v(t) = -16t^2 + v(0)t + h(0)$

In which:

• v(0) is the initial velocity.

• h(0) is the initial height.

For Peter, we have that:

$h(t) = -16t^2 + 64t + 80$

Which is a quadratic equation with coefficients $a = -16, b = 64, c = 80$

Considering that a < 0, the maximum value is:

$h_{MAX} = -\frac{\Delta}{4a} = -\frac{b^2 - 4ac}{4a}$

Hence:

$h_{MAX} = -\frac{64^2 - 4(-16)(80)}{4(-16)} = 144$

For Sophia:

• The initial height is half of Kevin, hence $h(0) = c = 40$.

• Twice the initial velocity, hence $v(0) = b = 128$

$h_{MAX} = -\frac{128^2 - 4(-16)(40)}{4(-16)} = 296$

296 - 144 = 152

The difference between the maximum heights is of 152 feet.

A similar problem is given at brainly.com/question/24713268