I am sooo lost on this math problem, please help?

Question

The equation; h = -16t(squared)+112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112 ft/s, where t is the time after the arrow leaves the ground. Find the time it takes the arrow to reach a height of 180 ft

Answer 1

Quadratic solution

-16t^2 +112*t -180 = 0
ax^2 + bx + c = 0

a = -16
b = 112
c = -180

t = (-b +- sqrt(b^2 - 4ac)) / 2a

t1 = (-112 + sqrt(112^2 - 4*16*180)) / (-2*16)
t2 = (-112 - sqrt(112^2 - 4*16*180)) / (-2*16)

t1 = 2.5 (on the way up)
t2 = 4.5 (on the way down)

Answer 2

Given the equation
h = -16t² + 112t
where h is the height and t is the time.

Since you need to find the time it takes the arrow to reach a height of 180 ft, you let the height h = 180 ft and then solve for t. Substituting the value of h,
180 = -16t² + 112t

You will easily see that there is a "t²" and a "t". Therefore your equation is a "2nd degree polynomial equation" or "equation of a parabola" or "quadratic equation." Solving this type of problem requires you to transform the equation into its general form ax² + bx + c = 0 (meaning you transpose everything to the left)
16t² - 112t + 180 = 0

To simplify, divide everything by 4:
4t² - 28t + 45 = 0

So there is your equation. If you already know how to solve this kind of equation, you will notice that it factors to
(2t - 9)(2t - 5) = 0

And the solutions are t = 9/2 and t = 5/2

But if you don't know yet, then here it is:

Methods of solving quadratic equation:
1) Factoring
The fastest, yet not always effective method is factoring. If you already know how to factor, then try to factor everything out. For example:
x² - 3x + 2 = 0
You can factor it as:
(x - 1)(x - 2) = 0
then you set all factors to equal zero:
x - 1 = 0 or x - 2 = 0
x = 1 or x = 2
so you get the solutions for the equation.

2) Completing the square
If you do not know how to factor or the equation is too hard to factor, you can use this method. It follows very easy steps:
take x² - 3x + 2 = 0 for example:
Transpose the constant
x² - 3x = -2
Divide by the coefficient of x². (since in this example, the coef. of x² is 1, there is no need to divide)
x² - 3x = -2
Take the coef. of x, divide it by 2, then square, then add it to both sides of the equation:
x² - 3x + (-3/2)² = -2 + (-3/2)²
x² - 3x + 9/4 = -2 + 9/4
Since the left part is a perfect square trinomial, factor it. Also simplify the right part.
(x - 3/2)² = 1/4
Get the square root of both sides
x - 3/2 = ± 1/2
Finally transpose the constant on the left side:
x = 3/2 ± 1/2
x = 3/2 + 1/2 or x = 3/2 - 1/2
x = 4/2 or x = 2/2
x = 2 or x = 1.

You see that we get the same solutions:
3) Quadratic equation:
You can solve for the general form ax² + bx + c = 0 by completing the square:
ax² + bx + c = 0

ax² + bx = -c

x² + b/a x = -c/a

x² + b/a x (b/2a)² = -c/a + (b/2a)²

x² + b/a x + b²/4a² = -c/a + b²/4a²

(x + b/2a)² = (-4ac + b²)/4a²

(x + b/2a)² = (b² - 4ac)/4a²

x + b/2a = ±√(b² - 4ac)/2a

x = -b/2a ± √(b² - 4ac)/2a

x = [-b ± √(b² - 4ac)]/2a

You can use this formula to solve for the values of x. Take x² - 3x + 2 = 0 for example. Here a = 1,b = -3 and c = 2. Substituting to the formula:
x = {-b ± √[b² - 4ac]}/2a
x = {-(-3) ± √[(-3)² - 4(1)(2)]}/2(1)
x = [3 ± √(9 - 8)]/2
x = [3 ± √(1)]/2
x = (3 ± 1)/2
x = (3 + 1)/2 or x = (3 - 1)/2
x = (4)/2 or x = (2)/2
x = 2 or x = 1

Now let's go back to the original problem:
4t² - 28t + 45 = 0
1) factoring:
It is factored as
(2t - 9)(2t - 5) = 0
Set each factor equal to zero
2t - 9 = 0 or 2t - 5 = 0
2t = 9 or 2t = 5
t = 9/2 or t = 5/2

2)completing the square:
4t² - 28t + 45 = 0
4t² - 28t = -45
t² - 7t = -45/4
t² - 7t + (-7/2)² = -45/4 + (-7/2)²
t² - 7t + 49/4 = -45/4 + 49/4
(t - 7/2)² = 4/4
(t - 7/2)² = 1
t - 7/2 = ±1
t = 7/2 ± 1
t = 7/2 + 1 or t = 7/2 - 1
t = 9/2 or t = 5/2

3)quadratic formula
4t² - 28t + 45 = 0
ax² + bx + c = 0
Here,
x = t
a = 4
b = -28
c = 45
Thus,
x = [-b ± √(b² - 4ac)]/2a
t = {-(-28) ± √[(-28)² - 4(4)(45)]}/2(4)
t = [28 ± √(784 - 720)]/8
t = [28 ± √64]/8
t = (28 ± 8)/8
t = (28 + 8)/8 or t = (28 - 8)/8
t = 36/8 or t = 20/8
t = 9/2 or t = 5/2

We notice that all methods yield to
t = 9/2 or t = 5/2

The question asks for the TIME IT TAKES for the arrow to reach the height of 180 ft, so you take the smaller value
t = 5/2 s.

^_^

Answer 3

h = -16t^2 + 112t

The 112 is the initial velocity (ft/s)
-16 is the accleration due to gravity (ft/s^2)
h is the height they give you, 180 ft

Plug in "h"

180 = -16t^2 + 112t
16t^2 - 112t + 180 = 0 ----> rearrange the terms

4(4t^2 - 28t + 45) = 0 ---> factor out a "4" to simplify the calculation

(4t^2 - 28t + 45) = 0

t = [ -(-28) +/- sqrt( (-28^2) - (4)(4)(45) ) ] / (2)(4)
= [ 28 +/- sqrt( 784 - 720 ) ] / 8
= [ 28 +/- sqrt( 64 ) ] / 8
= [ 28 +/- 8 ] / 8

t = (36/8) and (20/8)
2.5 seconds and 4.5 seconds

This means the arrow reaches that height twice. It first hits that height at 2.5 seconds. It then hits it's maximum height and begins to descend back to earth, going through the height of 180 ft again at 4.5 seconds.

Answer 4

The equation you have is thus:

h = -16t^2 + 112t
(^2 means to the power of two, or "squared")

To find the time at which it reaches 180 feet, you just substitute. You know that the height is 180 feet, therefore for h, we will sub in 180 feet:

180' = -16t^2 + 112t

Now we have an equation with one variable (t), so we need to solve for t. The first step is to equate everything to zero, by "moving" it all to one side:

180' = -16t^2 + 112t
16t^2 + 180' = 112t [by adding 16t^2 to both sides]
16t^2 - 112t + 180' = 0 [by subtracting 112t from both sides]

Now we either factor this, or use the quadratic equation. Since the factors of this equation are not easily obvious, we'll use the quadratic equation:

t = [-b +/- sqrt(b^2 - 4ac)]/[2a]
where a = 16, b = -112 and c = 180
(note sqrt denotes a square root)

Now sub in a, b and c:
t = [-(-112) +/- sqrt((-112)^2 - 4(16)(180))]/[2(16)]
t = [112 +/- sqrt(1024)]/32
t = [112 + 32]/32 OR t = [112 - 32]/32
t = 4.5s OR t = 2.5s

So our answers for t are both 2.5 seconds and 4.5 seconds. This means that 2.5 seconds from releasing the arrow, it reaches 180 feet. It then continues upward until it reaches its maximum height, then fall back to the ground, reaching 180 feet at 4.5 seconds.

Hope this helps.

Answer 5

Given:
h = -16t² + 112t and h = 180 ft/s

h = -16t² + 112t
180 = -16t² + 112t
+16t² - 112t + 180 = 0 (÷ 4)
4t² - 28t + 45 = 0

Use the formula: [-b±√(b² - 4ac)] / 2a
[-(-28)±√(-28)² - 4.4.45] / 2(4)

[28 ± √(784 - 720)] / 8

(28 ± 8) / 8

t = 4∙5 sec and t = 2∙5 sec.

Answer 6

180=-16t^2+112t
or 16t^2-112t=-180
or 16t^2-112t+180=0
or t^2-7t+11.25=0
Hence, t=(7(+-)sqrt(7^2-(4*1*11.25)))/(2*1)
or t=(7(+-)sqrt(4))/2
or t=(7(+/-)2)/2
Hence t=9/2 for (7+2)/2 which is the time taken after it reaches the maximum height and comes back down.
And t=5/2 for (7-2)/2 which is the time taken by the particle when it is moving upwards i.e just after thrown upwards from the ground.

Answer 7

if h equals 180, and
if h equals -16 * t* t +112*t
what does t equal

180 = -16 * t * t +112 *t
or
-16 * t * t +112 *t -180 = 0

This is a quadratic. Its simple to find the roots of this, you should know that formula.

Answer 8

set h=180, solve for "t"

180 = -16t^2 +112t

16t^2 -112t +180 = 0
8 56 90
4 28 45

4t^2 - 28t +45 = 0

t = "(-b plusorminus sqrt (b^2 - 4ac)) / 2a "

a=4
b=-28
c=45


"do the math"

Answer 9

just set h to 118feet and solve the resuting equation.

h = -16t(squared)+112t

118 = -16t(squared)+112t <=== solve this.

Answer 10

a man is carrying 100 coconuts by carts.he has to cross 100 check posts and at every checkpost he has to give 1coconut.inspite of this the man crosses all the checkposts and saves a few coconuts which he sells in the market and earns good money and returns homehappily.How many coconuts he saved?
if you get the answer think you have solved your math