Respuesta :
Answer:
When do you hit the water?
1.075 seconds after you jump.
What is your maximum height?
the maximum height is 12.5626 ft
Step-by-step explanation:
The equation:
h(t) = -16*t^2 + 6*t + 12
Is the height as a function of time.
We know that the initial height is the height when t = 0s
h(0s) = 12
and we know that the diving board is 12 foot tall.
Then the zero in h(t)
h(t) = 0
Represents the surface of the water.
When do you hit the water?
Here we just need to find the value of t such that:
h(t) = 0 = -16*t^2 + 6*t + 12
Using the Bhaskara's formula, we get:
[tex]t = \frac{-6 \pm \sqrt{6^2 - 4*(-16)*12} }{2*(-16)} = \frac{-6 \pm 28.4}{-32}[/tex]
Then we have two solutions, and we only care for the positive solution (because the negative time happens before the jump, so that solution can be discarded)
The positive solution is:
t = (-6 - 28.4)/-32 = 1.075
So you hit the water 1.075 seconds after you jump.
What is your maximum height?
The height equation is a quadratic equation with a negative leading coefficient, then the maximum of this parabola is at the vertex.
We know that the vertex of a general quadratic:
a*x^2 + b*x + c
is at
x = -b/2a
Then in the case of our equation:
h(t) = -16*t^2 + 6*t + 12
The vertex is at:
t = -6/(2*-16) = 6/32 = 0.1875
Evaluating the height equation in that time will give us the maximum height, which is:
h(0.1875) = -16*(0.1875 )^2 + 6*(0.1875) + 12 = 12.5626
And the height is in feet, then the maximum height is 12.5626 ft
You hit the water 1.075 seconds after you jump.
The maximum height is 12.56.
Given that,
You are jumping off the 12-foot diving board at the municipal pool.
You bounce up at 6 feet per second and drop to the water your height h (in feet) above the water in terms of t seconds is given by,
[tex]\rm h(t)=-16t^2+6t+12[/tex]
We have to determine,
When do you hit the water?
What is your maximum height?
According to the question,
You bounce up at 6 feet per second and drop to the water your height h (in feet) above the water in terms of t seconds is given by,
[tex]\rm h(t)=-16t^2+6t+12[/tex]
1. The initial height is the height when t = 0 second,
[tex]\\\rm h(t)=-16t^2+6t+12\\\\\rm h(0)=-16(0)^2+6(0)+12\\\\ h(0) = 12[/tex]
And the diving board is 12 feet tall.
Then the zero in h(t)
h(t) = 0
When h(t) = 0 represent the surface of the water,
[tex]\rm h(t)=-16t^2+6t+12\\\\\rm -16t^2+6t+12 =0\\\\8t^2-3t-6=0\\\\x = \dfrac{-(-3)\pm \sqrt{(-3)^2-4\times8\times(-6)}}{2\times 8}}\\\\x = \dfrac{3\pm \sqrt{9+192}}{16}}\\\\x = \dfrac{3\pm \sqrt{201}}{16}}\\\\x = \dfrac{3\pm 14.77}{16}}\\\\ x = \dfrac{3+14.77}{16} \ and \ x = \dfrac{3-14.77}{16} \\\\x = \dfrac{17.77}{16} \ and \ x = \dfrac{-11.77}{16} \ \\\\ x = 1.075 \ and \ x = -0.855[/tex]
The value of x can not be negative then the value of x is 1.075.
Therefore, you hit the water 1.075 seconds after you jump.
2. The maximum height is reached by you, you first derivative the function h(t) with respect to t:
[tex]\rm \dfrac{dh(t)}{dt} = \dfrac{d(-16t^2+6t+12)}{dt}\\\\\rm \dfrac{dh(t)}{dt} = \dfrac{d(-16t^2)}{dt} +\dfrac{d(6t)}{dt}+\dfrac{d(12)}{dt}\\\\\rm \dfrac{dh(t)}{dt} = 2\times (-16t) + 6\times 1+0\\\\ \dfrac{dh(t)}{dt} = -32t+6\\\\[/tex]
To find the maximum height the value first derivative is equal to zero.
[tex]\rm \dfrac{dh(t)}{dt} = 0\\\\-32t+6=0\\\\-32t=-6\\\\t = \dfrac{6}{32}\\\\t = \dfrac{3}{16}\\[/tex]
Substitute the value of t in the equation to find the maximum height,
[tex]\rm h(\dfrac{3}{16})=-16(\dfrac{3}{16})^2+6(\dfrac{3}{16})+12\\\\ h(\dfrac{3}{16})=-(\dfrac{9}{16})+6(\dfrac{3}{16})+12\\\\ h(\dfrac{3}{16})=-\dfrac{-9}{16}+\dfrac{18}{16}+12\\\\ h(\dfrac{3}{16})= \dfrac{-9+18+192}{16} \\\\ h(\dfrac{3}{16}) = \dfrac{201}{16}\\\\h(\dfrac{3}{16}) = 12.56[/tex]
Hence, The maximum height is 12.56.
For more details refer to the link given below.
https://brainly.com/question/20120328
