For purposes of completing our calculations, we're going to assume that
the experiment takes place on or near the surface of the Earth.
The acceleration of gravity on Earth is about 9.8 m/s², directed toward the
center of the planet. That means that the downward speed of a falling object
increases by 9.8 m/s for every second that it falls.
3 seconds after being dropped, a stone is falling at (3 x 9.8) = 29.4 m/s.
That's the vertical component of its velocity. The horizontal component is
the same as it was at the instant of the drop, provided there is no horizontal
force on the stone during its fall.
Mitochondria breaks down sugar to release energy to the cell.
-- The long line and short line close together at the left side
of the diagram represent a single-cell battery.
It's the only one in this diagram.
It's a device that stores chemical energy and delivers it on demand.
-- The zig-zag lines with circles around them represent light bulbs.
There are three of them in this diagram.
They are devices used to produce light by dissipating electrical energy.
-- The zig-zag lines without circles, at the top of the diagram,
represent resistors.
There are two of them in this diagram.
They are devices used to change or control electrical parameters
within a circuit by dissipating electrical energy.
-- The short straight line between two small circles at the bottom
of the diagram represents a switch.
There is only one switch in this circuit.
It's a device used to easily and quickly start or stop the flow of current
past a certain point in a circuit.
In this circuit ...
-- When the switch is closed (as drawn), the light bulb nearest the battery
glows brightest, the light bulb in the middle glows less bright, and the light
bulb on the right side glows dimmest of all.
-- When the switch is open, the light bulb nearest the battery glows, and
neither of the other two light bulbs glows at all.
Answer:
just before landing the ground
Explanation:
Let the velocity of projection is u and the angle of projection is 30°.
Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.
initial horizontal component of velocity, ux = u Cos 30
initial vertical component of velocity, uy = u Sin 30
Time of flight is given by
Final horizontal component of velocity, vx = ux = u Cos 30
Let vy is teh final vertical component of velocity.
Use first equation of motion
vy = uy - gT
vy = - u Sin 30
The magnitude of final velocity is given by
v = u
Thus, the velocity is same as it just reaches the ground.
