All categories

3 years ago

A ball is thrown from a height of

1 answer:

katrin2010 [14]3 years ago

3 0

Formula is: h = vi · t + 1/2 g · t²
For the final position ( on the ground ): h = - 255 ft
vi = - 21 ft/s, g = -9.81 m/s² = - 32.174 ft/s²
- 255 = - 21 t - 1/2 · 32.174 t²
16.087 t² + 21 t - 255 = 0
t 1/2 = ( - 21 +/- √ (441 + 16,408.74 ))/ 32.174 =
= ( -21+/- 129.8 ) / 32.174 = 108.8 / 32.174 = 3.38 s
Answer: The ball hits the ground after 3.38 s.

You might be interested in

How to make 135by 210in lowest term

Ainat [17]

Answer:

9/14

Step-by-step explanation:

.................

5 0

3 years ago

Read 2 more answers

How do i do this equation <br> -3(-2y-4)-5y-2=

sp2606 [1]

Answer:

combined like terms and then follow the order of operations.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What if i want my own room but there are six kids and only three rooms then what do i do

ANEK [815]

Answer:

put all the kids in a room with the parents and have the other room to yourself

Step-by-step explanation:

6 0

3 years ago

W-9=15 solve this equation and show all steps

jarptica [38.1K]

W= 24
Because use inverse operations so add 9 on both sides, that means the -9 on the left side is canceled out, remaining with W=24

8 0

3 years ago

Read 2 more answers

Rectangle A is a scale drawing of Rectangle B and has 25% of its area. If rectangle A has side lengths of 4 cm and 5 cm , what a

SpyIntel [72]

Answer:

8 cm and 10 cm

Step-by-step explanation:

Hello, <em> </em>I can help you with this.

Step 1

According to the question there are two rectangles A and B,

Rectangle A is a scale drawing of Rectangle B and has 25% of its area

in other words

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\$

Step 2

Let

Rectangle A

length (1)= 4 cm

length (2)= 5 cm

$Area_{A}=4\ cm * 5\ cm\\Area_{A}=20\ cm^{2}$

put this value into equation 1

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\\\20\ cm^{2} =0.25*Area_{B} \\divide\ each\ side\ by\ 0.25\\\frac{20\ cm^{2} }{0.25}=\frac{0.25}{0.25}*Area_{B}\\ Area_{B}=80\ cm^{2}$

Now, we know the area of rectangle B, to know its length we need to formule other equation

Step 3

$Area_{B}=80\ cm^{2}\\length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\$

the ratio between the lengths must be constant, so the ratio of A must be equal to ratio in B, then

$\frac{length(1A)}{length(2A)}=\frac{length(1B)}{length(2B)} \\\\\\frac{4}{5}= \frac{length(1B)}{length(2B)}\\0.8=\frac{length(1B)}{length(2B)}\\length(1B)=0.8*length(2B) (Equation 3)$

Step three

using Eq 1 and Eq 2 find the lengths

put the value of length(1B) into equation (2)

$length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\\(0.8*length(2B)) (*length (2B)=80\ cm^{2} \\\\0.8*(length (2B))^{2} =80\ cm^{2}\\(length (2B))^{2} =\frac{80\ cm^{2}}{0.8} \\(length (2B))^{2}=100\\\sqrt{(length (2B))^{2}}=\sqrt{100\ cm^{2}} \\ length (2B)=10\ cm$

Now, put the value of length(2B) into equation 3 to know length (1B)

$length(1B)=0.8*length(2B)\\length(1B)=0.8*10\ cm\\length(1B)=8 cm$

I really hope this helps you, have a great day.

6 0

3 years ago