rjuan00046
Answered

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 :

merlynthewhizz
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

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

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

[tex]0=-16 t^{2}+224t [/tex]
[tex]0=-16t(t-14)[/tex]
[tex]-16t=0[/tex] and [tex]t-14=0[/tex]
[tex]t=0[/tex] and [tex]t=14[/tex]

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




View image merlynthewhizz
johanrusli

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:

[tex]ax^2 + bx + c = 0[/tex]

The formula to solve this equation is :

[tex]\large {\boxed {x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} } }[/tex]

Let's try to solve the problem now.

Question A:

Given :

[tex]h = -16 t^2 + 224t[/tex]

The rocket reaches 720 feet → h = 720 feet

[tex]720 = -16 t^2 + 224t[/tex]

[tex]16 t^2 - 224t + 720 = 0[/tex]

[tex]16 (t^2 - 14t + 45 = 0)[/tex]

[tex]t^2 - 14t + 45 = 0[/tex]

[tex]t^2 - 9t - 5t + 45 = 0[/tex]

[tex]t(t - 9) - 5(t - 9) = 0[/tex]

[tex](t - 5)(t - 9) = 0[/tex]

[tex]t = 5 ~ or ~ t = 9[/tex]

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

Question B:

The rocket hits the ground → h = 0 feet

[tex]0 = -16 t^2 + 224t[/tex]

[tex]16 (t^2 - 14t ) = 0[/tex]

[tex]t^2 - 14t = 0[/tex]

[tex]t( t - 14 ) = 0[/tex]

[tex]t = 0 ~ or ~ t = 14[/tex]

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

Learn more

  • method for solving a quadratic equation : https://brainly.com/question/10278062
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  • best way to solve quadratic equation : https://brainly.com/question/9438071

Answer details

Grade: College

Subject: Mathematics

Chapter: Quadratic Equation

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

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