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3 years ago

To catch a ball, a professional baseball player leaps into the air with an initial velocity of

Mathematics

1 answer:

REY [17]3 years ago

4 0

Answer:

The answer is below

Step-by-step explanation:

a) The baseball player leaps into the air with an initial velocity of 14 feet per second. This means that the player leaps from the ground, the initial height is therefore zero.

The height (y) is given by the formula:

y(t) = ut - (1/2)gt² + initial heigth

u = initial velocity = 14 ft/s, t = time taken, g = acceleration due to gravity = 32 ft/s². Substituting:

y(t) = 14t - (1/2) * 32 *t² + 0

y(t) = 14t - 16t²

b) when the player is on the ground, the height = 0. hence:

0 = 14t - 16t²

16t² - 14t = 0

t(16t - 14) = 0

t = 0 or 16t - 14 = 0

t = 0 or 16t = 14

t =0 or t = 0.875

Hence t = 0.875 seconds

c) The domain is the set of possible values for the time. Hence:

Domain = (0, 0.875] = 0 < t ≤ 0.875

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3 years ago

I am having trouble with this relative minimum of this equation.<br>

Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

$y=4x^3+13x^2-12x-56$

$y'=(4x^3+13x^2-12x-56)'$

$y'=(4x^3)'+(13x^2)'-(12x)'-(56)'$

$y'=4(x^3)'+13(x^2)'-12(x)'-0$

$y'=4(3x^2)+13(2x^1)-12(1)$

$y'=12x^2+26x-12$

Now we set y' equal to 0 and solve for the critical numbers.

$12x^2+26x-12=0$

Divide both sides by 2:

$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

$b=13$

$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

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

$x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}$

$x=\frac{-13 \pm \sqrt{169+144}}{12}$

$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

$f(x)=4x^3+13x^2-12x-56$