davis52202
24.11.2021 •
Physics
If the near point of your eye is at 75 cm, you are
Solved
Show answers
More tips
- L Legal consultation Juvenile Justice: Who Needs It?...
- F Family and Home How to Choose the Best Diapers for Your Baby?...
- F Family and Home Parquet or laminate, which is better?...
- L Leisure and Entertainment How to Properly Wind Fishing Line onto a Reel?...
- L Leisure and Entertainment How to Make a Paper Boat in Simple Steps...
- T Travel and tourism Maldives Adventures: What is the Best Season to Visit the Luxurious Beaches?...
- H Health and Medicine Kinesiology: What is it and How Does it Work?...
- O Other How to Choose the Best Answer to Your Question on The Grand Question ?...
- L Leisure and Entertainment History of International Women s Day: When Did the Celebration of March 8th Begin?...
- S Style and Beauty Intimate Haircut: The Reasons, Popularity, and Risks...
Answers on questions: Physics
- P Physics What is the push and pull of a man about to hit a golf ball with a golf club. is it causing a change in speed, or direction, or both?...
- S Social Studies Information which we believe to be true and for which we have justification or evidence: answer a. knowledge b. induction c. deduction d. emotion...
- M Mathematics Which value for the constant c makes z = 4 an extraneous solution in the following equation?...
- M Mathematics In the figure below, the vertices of triangle RST are on a circle. 8 in. 6 in. s • Line segment TS contains the center of the circle. • The perimeter of triangle...
- H History What was khrushchev s view of u.s. nations? according to khrushchev, what would happen?...
- P Physics How long will it take a shell fired from a cliff at an initial velocity of 800 m/s at an angle 30 degrees below the horizontal to reach the ground 150m below?...
Ответ:
I will answer in English.
Ok, we know that a ball is trowed up, and the maximum height is 21.4m
If we suppose that the initial height of the ball is h = 0 (this means that is trowed from the ground) the equations of movement will be:
for the acceleration we only have the gravitational acceleration, so we have:
a(t) = -9.8m/s^2
for the velocity we can integrate over time and get:
v(t) = (-9.8m/s^2)*t + v0
where v0 is the initial speed.
For the position we integrate over the time again, and as the initial position is 0m, here we do not have any integration constant.
p(t) = (1/2)(-9.8m/s^2)*t^2 + v0*t
a) the maximum height is reached when the velocity is equal to zero, this happens at the time:
(-9.8m/s^2)*t + v0 = 0
t = v0/(9.8m/s^2)
now we can replace this time in the equation for the position:
p = 21.4m = (-4.9m/s^2)*(v0/(9.8m/s^2))^2 + v0^2/(9.8m/s^2)
and solve it for v0, i will stop writing the units so it is easier to read:
21.4 = -4.9*(v0/9.8)^2 + v0^2/9.8
21.4 = v0^2*(0.05)
v0 = √(21.4/0.05) = 20.7
so the initial speed is 20.7 m/s.
b) the things i supposed are:
The initial position is p(0) = 0
there are no things like air resistance or wind.
c) we can take t = 3.05s and put this in the position equation:
p(3.05s) = (1/2)(-9.8m/s^2)*(3.05s)^2 + 20.7m/s*3.05s = 17.5m
3.05 seconds after the launch, the ball will be 17.5 m above the ground.
d) we put t = 3.05s in the velocity equation:
v(3.05s) = (-9.8m/s^2)*3.05s + 20.7m/s = -9.19m/s
which means that now the ball is falling down.