Question

A rocket is launched straight up from the ground, with an initial velocity of 224 feet per second. The equation for the height of the rocket at time t is given by:
h=-16t^2+224t

(Use quadratic equation)

A.) Find the time when the rocket reaches 720 feet.


B.) Find the time when the rocket completes its trajectory and hits the ground.

Answer 1

A.) The rocket reaches 720 feet in 5 seconds and 9 seconds.

B.) The rocket completes its trajectory and hits the ground in 14 seconds

Further explanation

A quadratic equation has the following general form:

$ax^2 + bx + c = 0$

The formula to solve this equation is :

$\large {\boxed {x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} } }$

Let's try to solve the problem now.

Question A:

Given :

$h = -16 t^2 + 224t$

The rocket reaches 720 feet → h = 720 feet

$720 = -16 t^2 + 224t$

$16 t^2 - 224t + 720 = 0$

$16 (t^2 - 14t + 45 = 0)$

$t^2 - 14t + 45 = 0$

$t^2 - 9t - 5t + 45 = 0$

$t(t - 9) - 5(t - 9) = 0$

$(t - 5)(t - 9) = 0$

$t = 5 ~ or ~ t = 9$

The rocket reaches 720 feet in 5 seconds and 9 seconds.

Question B:

The rocket hits the ground → h = 0 feet

$0 = -16 t^2 + 224t$

$16 (t^2 - 14t ) = 0$

$t^2 - 14t = 0$

$t( t - 14 ) = 0$

$t = 0 ~ or ~ t = 14$

The rocket completes its trajectory and hits the ground in 14 seconds

Learn more

• method for solving a quadratic equation : brainly.com/question/10278062
• solution(s) to the equation : brainly.com/question/4372455
• best way to solve quadratic equation : brainly.com/question/9438071

Answer details

Grade: College

Subject: Mathematics

Chapter: Quadratic Equation

Keywords: Quadratic , Equation , Formula , Rocket , Maximum , Minimum , Time , Trajectory , Ground

Answer 2

We can model the equation of the height of the rocket as ∩-shape curve as shown below

Part A: 

The time when the height is 720 feet

$720 = -16 t^{2}+224t$, rearrange to make one side is zero
$16 t^{2}-224t+720=0$, divide each term by 16
$t^{2} -14t+45 =0$, factorise to give
$(t-9)(t-5)=0$
$t=9$ and $t=5$

So the rocket reaches the height of 720 feet twice; when t=5 and t=9

Part B:

We will need to find the values of t when the rocket on the ground. The first value of t will be zero as this will be when t=0. We can find the other value of t by equating the function by 0

$0=-16 t^{2}+224t$
$0=-16t(t-14)$
$-16t=0$ and $t-14=0$
$t=0$ and $t=14$

So the time interval when the rocket was launched and when it hits the ground is 14-0 = 14 seconds