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VMariaS [17]
3 years ago
14

Type the correct answer in the box. Farmer Jim is building a new chicken coop in the shape of a rectangular prism. He needs the

height of the coop to be 2 feet shorter than the width, and needs the length of the coop to be 10 feet longer than the width. Create an equation to represent the volume of the chicken coop, y, in terms of the width, x, in feet. Enter the equation in the field by replacing the values of a, b, c, and d. Big fraction Parentheses Vertical bars Square root Root Superscript (Ctrl+Up) Subscript (Ctrl+Down) Plus sign Minus sign Middle dot Multiplication sign Equals sign Less-than sign Greater-than sign Less-than or equal to Greater-than or equal to Pi Alpha Beta Epsilon Theta Lambda Mu Rho Phi Sine Cosine Tangent Arcsine Arccosine Arctangent Cosecant Secant Cotangent Logarithm Logarithm to base n Natural logarithm Bar accent Right left arrow with under script Right arrow with under script Angle Triangle Parallel to Perpendicular Approximately equal to Tilde operator Degree sign Intersection Union Summation with under and over scripts Matrix with square brackets xy = ax3+bx2+cx+d
Mathematics
1 answer:
Digiron [165]3 years ago
4 0

Answer:

20 feet

Step-by-step explanation:

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An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object
Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2  + \frac{1}{2} at^2\\3) v = v_0 + at

What do we already know?

  • v_0 = 29.4 ms^-1
  • The launch is straight up
  • a = -9.81 ms^-2 this is the gravitational acceleration g
  • s_0 = 0 m, since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, v = 0 ms^-2

v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2

this is the equation of the whole trajectory that object makes.

and if you solve this by making s = 0, you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

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