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a_sh-v [17]
3 years ago
8

A basketball is thrown upwards. The height f(t), in feet, of the basketball at time t, in seconds, is given by the following fun

ction:
f(t) = -16t^2 + 44t + 12

Which of the following is a reasonable domain of the graph of the function when the basketball falls from its maximum height to the ground?

Physics
2 answers:
snow_lady [41]3 years ago
8 0

The correct answer would be  1.375 < t < 3  i hope this helps anyone


Vikki [24]3 years ago
3 0

Answer:

Domain is 1.375 s to 3 s

Explanation:

It is given that, A basketball is thrown upwards. The height f(t), in feet, of the basketball at time t, in seconds, is given by the following function:

f(t)=-16t^2+44t+12

We have to find the domain of the graph of the function when the basketball falls from its maximum height to the ground. It is shown in the attached graph.

It is clear that, the domain of the graph is from 1.375 s to 3 s. Its maximum height is 42.25 feet.

Hence, this is the required solution.      

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3 years ago
A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 48 ft above the ground. If the elevator can
belka [17]

Answer: 21.91 s

Explanation:

Given that,

Maximum height of the car, h = 48 ft

Acceleration of the elevator, a = 0.6 ft/s²

Deceleration of the elevator, -a = 0.3 ft/s²

Maximum speed of the elevator, v = 8 ft/s

Initial speed of the elevator, u = 0

If when the elevator accelerate from 0 to maximum velocity, v.

Let s be the vertical distance traveled during acceleration.

v² = u² - 2as

s = (v² - u²) / 2a

s = (8² - 0) / 2*0.6

s = 64 / 1.2

s = 53.33 ft

If when the elevator decelerates from maximum velocity, v to zero.

Let S be the vertical distance traveled during deceleration

u² = v² + 2aS

S = (u² - v²) / 2a

S = (0 - 8²) / 2 * 0.3

S = -64 / 0.6

S = 106.67 ft

Since he sum of s and S (i.e s + S) is greater than 48 ft, then the elevator will switch from acceleration to deceleration

without reaching the maximum velocity. Below, the switching point is labeled y.

v² = u² + 2ay

y = v²/2a

Inserting this into the earlier deceleration equation, we have

-v²/2 = d * [48 - (v²/2a)], where

d = deceleration

a = acceleration

Therefore, v = [4.√6. a √-(a.b/a)] / b

Where b = acceleration - deceleration

v = 4.382 ft/s

Using this newly found v, we proceed to find our s

s = (u² + v²)/2a

s = 19.2 / 1.2

s = 16 ft

The transport times for each segment are found from

v = u + a*t, thus upward t1

4.382 = 0 + 0.6 * t

t = 4.382/0.6

t = 7.303 s

Also,

4.382 = 0 + 0.3 * T

T = 4.382/0.3

T = 14.607 s

The total travel time is then t + T =

7.303 + 14.607

Total time of travel is 21.91 s

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