Ask question

Ask question

All categories

Nookie1986 [14]

3 years ago

11

Jake tosses a coin up in the air and lets it fall on the ground. The equation that

1 answer:

Phoenix [80]3 years ago

7 0

Answer:

$t = 1.71825$

Step-by-step explanation:

Given

$h(t) = -16t^2 + 24t + 6$

Required

When will the coin hit the ground

When the coin hits the ground, $h(t) = 0$

The expression $h(t) = -16t^2 + 24t + 6$ becomes

$0 = -16t^2 + 24t + 6$

Multiply through by -1

$16t^2 - 24t - 6 = 0$

Solve using quadratic formula

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = 16$

$b = -24$

$c = -6$

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

$t = \frac{-(-24)\±\sqrt{(-24)^2 - 4 *16 * -6}}{2 * 16}$

$t = \frac{24\±\sqrt{576 +384}}{32}$

$t = \frac{24\±\sqrt{960}}{32}$

$t = \frac{24\±30.984}{32}$

Split

$t = \frac{24+30.984}{32}$ or $t = \frac{24-30.984}{32}$

$t = \frac{54.984}{32}$ or $t = \frac{-6.984}{32}$

$t = 1.71825$ or $t = -0.21825$

But time can't be negative;

So:

<em>Time to hit the ground is 1.71825 seconds</em>

You might be interested in

-14+6b +7-2b=1+5b what is answer?~PLEASE SHOW WORK 45 POINTS

Over [174]

Answer:

b=-8

Step-by-step explanation:

Sorry I don't have time to show work right now and write it down but this is the answer.

8 0

2 years ago

Read 2 more answers

Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.

jarptica [38.1K]

$y''-y'+y=\sin x$

The corresponding homogeneous ODE has characteristic equation $r^2-r+1=0$ with roots at $r=\dfrac{1\pm\sqrt3}2$ , thus admitting the characteristic solution

$y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x$

For the particular solution, assume one of the form

$y_p=a\sin x+b\cos x$

${y_p}'=a\cos x-b\sin x$

${y_p}''=-a\sin x-b\cos x$

Substituting into the ODE gives

$(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x$

$-b\cos x+a\sin x=\sin x$

$\implies a=1,b=0$

Then the general solution to this ODE is

$\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}$

$y''-3y'+2y=e^x\sin x$

$\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2$

$\implies y_c=C_1e^x+C_2e^{2x}$

Assume a solution of the form

$y_p=e^x(a\sin x+b\cos x)$

${y_p}'=e^x((a+b)\cos x+(a-b)\sin x)$

${y_p}''=2e^x(a\cos x-b\sin x)$

Substituting into the ODE gives

$2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x$

$-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x$

$\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12$

so the solution is

$\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}$

$y''+y=x\cos(2x)$

$r^2+1=0\implies r=\pm i$

$\implies y_c=C_1\cos x+C_2\sin x$

Assume a solution of the form

$y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)$

${y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)$

Substituting into the ODE gives

$(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)$

$-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)$

$\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49$

so the solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}$

7 0

2 years ago

What is y when x = 60?

Virty [35]

Y can be any number with the given question

5 0

3 years ago

Read 2 more answers

A triangle has interior angles equal to 55, 70 and x. What is the value of X?

podryga [215]

The answer is C. 55

You get your answer by adding together the degrees of the angles you know 55+70

That equals 125 degrees and subtract that from 180 degrees

180-125= 55

8 0

2 years ago

A man is filling an empty 5-gallon bucket with water at a constant rate. there are 2.4 gallons of water in the bucket after fill

r-ruslan [8.4K]

Answer: The answer is 0.2

Step-by-step explanation: 2.4g / 12 minutes

0.2g/minutes

5 0

2 years ago