Answer:
Maximum height, h = 11.32 meters
Explanation:
It is given that,
The baseball is thrown directly upward at time, t = 0
Initial speed of the baseball, u = 14.9 m/s
Ignoring the resistance in this case and using a = g = 9.8 m/s²
We have to find the maximum height the ball reaches above where it leaves your hand. Let the maximum height is h. Using third equation of motion as :
At maximum height, v = 0
and a = -g = -9.8 m/s²
h = 11.32 meters
Hence, the maximum height of the baseball is 11.32 meters.
The distance from the surface where the is measured as the hydrodynamic boundary layer thickness. The local exterior velocity is the same as the speed.
<h3>What is velocity?</h3>
- Velocity is the direction at which an object is moving and serves as a measure of the rate at which its position is changing as seen from a specific point of view and as measured by a specific unit of time (for example, 60 km/h northbound).
- In kinematics, the area of classical mechanics that deals with the motion of bodies, velocity is a fundamental idea.
- A physical vector quantity called velocity must have both a magnitude and a direction in order to be defined.
- Speed is the scalar absolute value (magnitude) of velocity; it is a coherent derived unit whose quantity is measured in metres per second (m/s or ms1) in the SI (metric system).
<h3>What is speed?</h3>
- The speed of an object, also known as v in kinematics, is the size of the change in that object's position over time or the size of the change in that object's position per unit of time, making it a scalar quantity.
- The instantaneous speed is the upper limit of the average speed as the duration of the time interval gets closer to zero.
- The average speed of an object in a period of time is the distance traveled by the object divided by the duration of the interval.
- Velocity and speed are not the same thing.
Learn more about velocity here:
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Energy is never created and energy never is destroyed
Well, before we discuss that, I think we have to carefully understand
and agree on something. We have to be very clear about what we
mean by 'weight' ... is it what you feel, or is it the product of
(your mass) x (the acceleration of gravity where you are).
If you're on a space ship, then any time your engine is not burning,
you feel weightless. It doesn't matter where you are, or what body
you may be near. If you're not doing a burn, and the only force on
you is the force of gravity, then you don't feel any weight at all.
But of we say that your 'weight' is the product of
(your mass) times (the acceleration of gravity where you are),
then it depends on where you are, and whether you're close to
the Earth or closer to the moon. You may not feel it, but you're
going to have weight, and it's going to change during your trip
in space.
You know that the force of gravity depends on how far you are
from the body that's attracting you.
-- As you travel from the Earth to the moon, gravity will pull you
less and less toward Earth, and more and more toward the moon.
-- Your weight will get less and less, until you reach the point
in space where the gravitational attractions are equal in both
directions. That's about 24,000 miles before you reach the
moon ... about 90% of the way there. At that point, your weight
is really zero, because the pull toward the Earth and the pull toward
the moon are equal.
-- From there, the rest of the way to the moon, your weight will
start to grow again. It begins at zero at the 'magic point', and it
grows and grows until you reach the moon's surface. When
you're there, your weight has grown to about 1/6 of what you
weigh on Earth, and it won't get any bigger. If you weigh
120 pounds on Earth, then you weigh about 19.86 pounds on
the moon ... PLUS your space suit, boots, heater/air conditioner,
oxygen tank, radiation shielding, radio, and all the other stuff that
you need to survive on the moon for a few hours.
Distance covered by the squirrel to look for an acorn :
d = ( 3 m/s ) × 10 s = 30 m.
Time taken to eat an Acron is 5 seconds.
Time taken to cover distance of 30 m with 2 m/s speed is :
Therefore, total time take to get back to where he started is ( 10+5+15 ) = 30 s.
Hence, this is the required solution.
