50pts.

A ball is thrown vertically in the air with a velocity of 95 ft/s. The ball is at a height of 120 ft.

sattari [20]

Answer:

The ball is at a height of 120 feet after 1.8 and 4.1 seconds.

Step-by-step explanation:

The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."

Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

$y = y_{o} + v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$y_{o}$ - Initial height of the ball, measured in feet.

$v_{o}$ - Initial speed of the ball, measured in feet per second.

$g$ - Gravitational constant, equal to $-32.174\,\frac{ft}{s^{2}}$ .

$t$ - Time, measured in seconds.

Given that $y_{o} = 0\,ft$ , $v_{o} = 95\,\frac{ft}{s}$ , $g = -32.174\,\frac{ft}{s^{2}}$ and $y = 120\,ft$ , the following second-order polynomial is found:

$120\,ft = 0\,ft + \left(95\,\frac{ft}{s} \right)\cdot t +\frac{1}{2}\cdot \left(-32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}$

$-16.087\cdot t^{2} + 95\cdot t -120 =0$

The roots of this polynomial are, respectively:

$t_{1} \approx 4.075\,s$ and $t_{2} \approx 1.831\,s$ .

Both roots solutions are physically reasonable, since $t_{1}$ represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas $t_{2}$ represents the instant when the ball the same height after reaching maximum height.

In nutshell, the ball is at a height of 120 feet after 1.8 and 4.1 seconds.

8 0

3 years ago

Isabelle bought some stationery. 1/3 of them were pencils. 5/8 of the remainder were erasers and the rest were rulers. The cost

kobusy [5.1K]

Answer:

Isabelle spent $52.50 more on Erasers than rulers.

Step-by-step explanation:

Step 1

Find the quantity of each stationery bought in fraction

Let us represent the total fraction of what Isabelle bought as: x

1/3x = quantity of pencils bought

5/8 of the remainder = quantity of erasers bought

The remainder = x - 1/3x = 2/3x

5/8 of 2/3x = 5/8 × 2/3x = 5/12x

Hence, 5/12x = quantity of erasers bought

The rest is rulers

1 - (1/3 + 5/12)

1 - (4 + 5/12)

1 - 9/12

1 - 3/4

= 1/4x

Hence, 1/4 = quantity of rulers bought.

Step 2

We find the number of stationeries that Isabelle bought.

The cost of the stationery is

a Pencil: $1.20

an eraser: $1

a ruler: $0.50.

Total amount spent on stationery = 169.50

We have this equation

1.20 × 1/3x + 1 × 5/12x + 0.50 × 1/4x = 169.50

0.4x + 0.4166666667x + 0.125x =

169.50

0.9416666667x = 169.50

x = 169.50 /0.9416666667

x = 179.99999999

Approximately , the number of stationeries that Isabelle bought = x = 180

Step 3

Find the number and the amount spent on each stationery that Isabelle bought

a) i) Quantity of pencils bought = 1/3 x

x = 180

= 1/3 × 180

= 60 pencils

ii) Amount spent on pencils

If 1 pencil = $1.20

60 pencils =

60 × 1.20 = $72

b) i) Quantity of Erasers bought = 5/12x

x = 180

= 5/12 × 180

= 75 pencils

ii) Amount spent on Eraser

If 1 pencil = $1

75 pencils =

75 × 1 = $75

c) i) Quantity of rulers bought = 1/4 x

x = 180

= 1/4 × 180

= 45 pencils

ii) Amount spent on pencils

If 1 pencil = $0.50

45 pencils =

45 × 0.50 = $22.50

Step 4

We were asked to calculate how much more did she spend on erasers than rulers.

In step 3, the amount spent on Erasers = $75

the amount spent on ruler = $22.5

The difference in the amount spent = $75 - $22.5

= $52.5

Therefore, Isabelle spent $52.50 more on Erasers than rulers.

3 0

3 years ago

Round 10.054 To 2 decimal places

antoniya [11.8K]

The answer to "Round 10.054 To 2 decimal places" is 10.05

3 0

3 years ago

Eddie's Evergreens sells Christmas trees and wreaths. Trees cost 3 dollars more than 4 times the price of each wreath. It costs

kari74 [83]

Answer:

$W = 16$

$T = 67$

Step-by-step explanation:

Represent trees with T and wreaths with W

Given

$T = 3 + 4 * W$

$2 * W + T = 99$

Solving (a): Cost of W

Substitute 3 + 4 * W for T in the second equation

$2*W + 3 + 4 * W = 99$

$2W + 3 + 4 W = 99$

Collect Like Terms

$2W + 4 W = 99 - 3$

$6W = 96$

Divide through by 6

$W =\frac{96}{6}$

$W = 16$

Hence, each wreath costs $16

Solving (b): Cost of T

$T = 3 + 4 * W$

Substitute 16 for W

$T = 3 + 4 * 16$

$T = 3 + 64$

$T = 67$

Hence, each tree costs $67

3 0

2 years ago

This trapezoid-based right pyramid has a volume of 48m^3

aliya0001 [1]

Answer:

18

Step-by-step explanation:

Khan Academy

3 0

2 years ago