Can someone help with this

Arianna kicks a soccer ball off the ground and into the air with an initial velocity of 42 feet per second. Assume the starting

beks73 [17]

A=g

v=⌠g dt

v=gt+C, where C=v initial

v=gt+vi

h=⌠v dt

h=gt^2/2+vit+C, where C=h initial

h=gt^2/2+vit+hi

We are told that vi=42 ft/s, hi=0, and we know g≈-32 ft/s^2 so

h(t)=-16t^2+42t

The ball will hit the ground when h=0 so

-16t^2+42t=0

-2t(8t-21)=0, since t>0

8t-21=0

8t=21

t=21/8

t=2.625 sec

t≈2.6 sec (to nearest tenth of a second)

7 0

3 years ago

Read 2 more answers

I buy a 5 kg jar of sweets for £25. Then i divide the sweets into 125g packets and sell them for 99p each

kotegsom [21]

Answer:

is that the whole question? it looks unfinished to me-

3 0

3 years ago

Plz plz plz plz plz plz hlep

suter [353]

Answer:

C. 18:30

Step-by-step explanation:

Divide the y values (games played) by the x values (games won) to find your ratio. The ratio across each of these pairs is 1.666667...

Using the ratio provided, 18:30, divide 30 by 18.

30/18 = 1.666667...

It's a match!

Therefore, the answer is C. 18:30

6 0

4 years ago

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

$P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%$

Hence, a) 4.6798, and b) 19.8%.

7 0

3 years ago

Determine the discount between the point(-4,2) and the line 4y=3×+6

frozen [14]

Answer:

d = 14/5

Step-by-step explanation:

The point (-4,2) means that;

At x = -4, y = 2

Now general form of a linear equation is;

Ax + By + C = 0

We are given;

4y = 3x + 6

Rearranging to the form of a linear equation gives;

3x - 4y + 6 = 0

Thus, A = 3, B = -4 and C = 6

Thus, at point (-4,2), distance between them is;

d = (3(-4) - 4(2) + 6)/√(3² + (-4)²)

d = -14/5

We will take the absolute value.

Thus; d = 14/5

6 0

3 years ago