Answer:
47.4 m
Explanation:
When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.
In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is
Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation
where
s is the vertical displacement
u = 0 is the initial velocity
t = 3.11 s is the time
is the acceleration of gravity (taking downward as positive direction)
Solving the formula, we find
Answer:
An atom makes up almost everything; it is the building block of this universe. An atom is made out of protons, electrons, and neutrons. Protons are positively charged particles, Electrons are negatively charged particles, and Neutrons are neutral. The protons, and neutrons are found in the very center of the atom, which is called the nucleus. The electrons orbit the nucleus.
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Answer:
The time of the second observer at the end of the car motion is 27 s
Explanation:
Initial time of the first observer, t₁ = 262 s
final time of the first observer, t₂ = 375 s
The time of the car motion, t = t₂ - t₁
t = 375 s - 262 s
t = 113 s
Initial time of the second observer, t₁ = -86 s
final time of the second observer, t₂ = ?
The time of the car motion, t = t₂ - t₁
113 = t₂ - t₁
113 = t₂ - (-86)
113 = t₂ + 86
t₂ = 113 - 86
t₂ = 27 s
Therefore, the time of the second observer at the end of the car motion is 27 s
<span>The magnitude of an object's average velocity is equal to its
average speed during the same period of time, when the distance
it covers is equal to its displacement during that time.
This happens when the object travels in a straight line.</span>
Answer:
<em>The two balls pass each other at a height of 5.53 m</em>
<em>vf1=17.97 m/s</em>
<em>vf2=-5.96 m/s</em>
Explanation:
<u>Vertical Motion</u>
An object thrown from the ground at speed vo, is at a height y given by:
Where t is the time and 
Furthermore, an object dropped from a certain height h will fall a distance y, given by:
Thus, the height of this object above the ground is:
The question describes that ball 1 is dropped from a height of h=22 m. At the same time, ball 2 is thrown straight up with vo=12 m/s.
We want to find at what height both balls coincide. We'll do it by finding the time when it happens. We have written the equations for the height of both balls, we only have to equate them:
Simplifying:
Solving for t:
The height of ball 1 is:
H = 5.53 m
The height of ball 2 is:
y=5.53 m
As required, both heights are the same.
The speed of the first ball is:
vf1=17.97 m/s
The speed of the second ball is:
vf2=-5.96 m/s
This means the second ball is returning to the ground when both balls meet
