Mathematics : asked on jademckinziemea

06.02.2023

Find the time that the rocket will hit the ground to the nearest 100th of second

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27.01.2022, solved by verified expert

Ranjana Bhatt

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Find the time that the rocket will hit the ground, №15209688, 06.02.2023 13:10

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Mathematics

Arocket is launched from a tower. the height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. using this equation, find the time that the rocket will hit the ground, to the nearest 100th of second. equation: y=-16x^2+153x+98

Step-by-step answer

P Answered by PhD

$10.17\ seconds$

Step-by-step explanation:

Given the following Quadratic equation:

y=-16x^2+153x+98

The steps to solve it are:

1. Substitute y=0 into the equation:

0=-16x^2+153x+98

2. Use the Quadratic formula $x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$ .

In this case:

$a=-16\\b=153\\c=98$

Substituting values into the Quadratic equation, you get (Round the values to the nearest hundreth):

$x=\frac{-153\±\sqrt{153^2-4(-16)(98)}}{2(-16)}\\\\x_1=10.17\\x_2=-0.60$

3. The positive value is the time that the rocket will hit the ground.

Therefore, the rocket will hit the ground after $10.17\ seconds$ .

Mathematics

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the time that the rocket will hit the ground, to the nearest 100th of second. y=-16x^2+170x+61 please hurry will give brainlest and do not just take advantage of points

Step-by-step answer

P Answered by PhD

Given : The rocket is launched from a tower y=-16x^2+170x+61. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation.
To find : The time that the rocket will hit the ground ?
Solution :
When the rocket hit the ground i.e. height became zero y=0,
Equation is y=-16x^2+170x+61.
Substitute y=0,
-16x^2+170x+61=0.
Using quadratic formula, x=-b+-√b^2-4ac/2a;
x=-170+-√(170)^2-4(-16)(61)/2(-16);
x=-170+-√28885/-32;
x=170+√28885/-32, x=170-√28885/-32;
x=10.59, 0.03.
Answer: The time taken is 0.03 seconds.

Mathematics

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after jaunch, x in seconds, by the given equation. Using this equation, find the time that the rocket will hit the ground, to the nearest 100th of second. y = -16x^2 + 116x + 75

Step-by-step answer

P Answered by PhD

To find the time of the maximum height of the rocket differentiate the equation of the height with respect to the time and then equate the differentiation by 0 to find the time of the maximum height;
y is the height of the rocket after launch, x seconds
y=-16x^2+116x+75;
- Differentiate y with respect to x
y'=-16(2)x+116;
y'=-32x+116;
-Equate y'by 0
0=-32x+116;
-Add 32x to both sides
32x=116
-Divide both sides by 32
x=3.62 seconds.

Answer: 3.62 seconds.

Mathematics

Step-by-step answer

P Answered by PhD

To find the time of the maximum height of the rocket differentiate the equation of the height with respect to the time and then equate the differentiation by 0 to find the time of the maximum height;
y is the height of the rocket after launch, x seconds
y=-16x^2+213x+68;
- Differentiate y with respect to x
y'=-16(2)x+213;
y'=-32x+213;
-Equate y'by 0
0=-32x+213;
-Add 32x to both sides
32x=213
-Divide both sides by 32
x=6.65 seconds.

Answer: 6.65 seconds.

Mathematics

Step-by-step answer

P Answered by PhD

To find the time of the maximum height of the rocket differentiate the equation of the height with respect to the time and then equate the differentiation by 0 to find the time of the maximum height;
y is the height of the rocket after launch, x seconds
y=-16x^2+213x+68;
- Differentiate y with respect to x
y'=-16(2)x+213;
y'=-32x+213;
-Equate y'by 0
0=-32x+213;
-Add 32x to both sides
32x=213
-Divide both sides by 32
x=6.65 seconds.

Answer: 6.65 seconds.

Mathematics

Step-by-step answer

P Answered by PhD

Given

y = -16x^2+129x+119

When rocket hit the ground y = 0

-16x^2+129x+119 = 0

By comparing with ax^2+bx+c = 0

a = -16 , b = 129 and c= 119

Discriminant (D)= b^2 - 4ac

D = 129^2 - 4×(-16)×119

= 16641 + 7616 = 24257

By using the quadratic formula

1) x = (-b+√D)/2a

x = (-129 + √24257)/2(-16)

x = (-129+155.7465)/-32 = 26.7465/(-32)

x = -0.8358

2) x = (-b-√D)/2a

x = (-129 - √24257)/2(-16)

x = (-129 - 155.7465)/(-32) = -284.7465/(-32)

X = 8.898

Since x is time and time can't be in negative

So x = 8.898sec

Time taken by rocket to hit the ground = 8.898sec

Mathematics

Step-by-step answer

P Answered by Specialist

Strong?

Step-by-step explanation:

PLS HELP!!!

A rocket is launched from a tower. The height of the rocket, y in feet, is related to

Mathematics

Step-by-step answer

P Answered by PhD

t = 15.42 s

Step-by-step explanation:

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation.

y=-16x^2+242x+73

We ned to find the time when the rocket will the ground. For this, y = 0

-16x^2+242x+73=0

It is a quadratic equation whose solution is given by :

$x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}$

Here, a = -16, b = 242 and c = 73

$x=\dfrac{-b+ \sqrt{b^2-4ac} }{2a}, x=\dfrac{-b-\sqrt{b^2-4ac} }{2a}\\\\x=\dfrac{-242+ \sqrt{(242)^2-4(-16)(73)} }{2(-16)}, x=\dfrac{-242- \sqrt{(242)^2-4(-16)(73)} }{2(-16)}\\\\=-0.295\ s, 15.42\ s$