Mathematics, 06.10.2019 05:00 maheshwarlall

Arocket 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.

Answers: 1

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Answers

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

and

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)

and

and

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

Arocket is launched straight up from the ground, with an initial velocity of 224 feet per second. th

Arocket is launched straight up from the ground, with an initial velocity of 224 feet per second. th

Answer from: zoey130

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 moremethod for solving a quadratic equation : linksolution(s) to the equation : linkbest way to solve quadratic equation : linkAnswer details

Grade: College

Subject: Mathematics

Chapter: Quadratic Equation

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

Answer from: Quest

step-by-step explanation:

Answer from: Quest

Hello!

the answer is:

$\frac{3^{4} }{3^{1}}=3^{3}$

why?

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

$\frac{a^{m} }{a^{n} }=a^{m-n}$

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

$\frac{3^{4} }{3^{1}}$

then, calculating we have:

$\frac{3^{4} }{3^{1}}=3^{4-1}=3^{3}$

hence, the answer is:

$\frac{3^{4} }{3^{1}}=3^{3}$

have a nice day!

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