There's no such thing as "time of the downward velocity", but I think I get the sense
of your question.
If the effects of air resistance can be disregarded, then any object thrown upwards
spends half of its time rising, and the identical amount of time falling back to the
height of your hand when you let it go.
the tangential velocity is equal to the angular velocity multiplied by the radius the tangential velocity is equal to the angular velocity multiplied by the radius
It's equal to the change in velocity (final velocity - initial velocity).
you are still. motion is at rest.
Yea it is.
It means they are the same.
When the upward and downward forces on a falling object are equal, the object reaches terminal velocity. At terminal velocity, the object stops accelerating and falls at a constant speed.
As velocity never exceeds the velocity of light.... so i hope a man running with the velocity of light will not be able to throw a ball with any velocity.......... we may get the maximum n minimum velocity with which that can be thrown mathematically that we may get it to be zero................
The ball returns to the ground with increasing velocity due to acceleration due to gravity. At a point (terminal velocity) the ball maintains a constant velocity (due to air resistance) This occurs when the weight of the ball is equal to the viscous drag of the air (air resistance) and upthrust (weight of air displaced).
If it is thrown at an angle, at the top of its path, its vertical velocity will be zero, however its horizontal velocity will be the same as its initial horizontal velocity minus whatever loss in speed as a result of air friction at that point. We won't know what that is without more information.
The downward acceleration of a thrown object in projectile motion is constant and equal to the acceleration due to gravity, which is approximately 9.81 m/s^2 on Earth. This acceleration acts vertically downward and affects the vertical motion of the object while the horizontal motion remains unaffected.
At the highest point, the velocity of the ball is zero because it momentarily stops before falling back down. The acceleration of the ball at the highest point is equal to acceleration due to gravity, which is directed downward toward the center of the Earth and is approximately 9.81 m/s^2.
No. However, the statement is true provided that the vertical component of the launch velocity for the two motions are the same. You also require that both motions end at the same level and that the air resistance etc can be disregarded. The first of these may not always be valid in school exercises, the second is usually implicit.
The velocity of the ball 2 seconds before it reaches its highest point will be equal in magnitude but opposite in direction to the velocity it had when it was thrown upwards. This is due to the acceleration of gravity acting on the ball and gradually reducing its upward velocity until it comes to a stop at the highest point of its trajectory.
The acceleration of a ball at the top of its trajectory when thrown straight upward is equal to the acceleration due to gravity, which is approximately 9.81 m/s^2 downward. At the top of its trajectory, the ball momentarily comes to a stop before reversing direction and accelerating downward.
The average acceleration of the object will be negative since it changes direction from upward to downward velocity. The acceleration will be constant because the object undergoes constant acceleration throughout the motion.
Yes, if it reaches terminal velocity, which is a constant velocity. When terminal velocity is reached, the downward gravitational force is equal to the upward force of air resistance, and the object no longer accelerates.
Skydivers reach terminal velocity because as they fall, the force of gravity pulling them downward is balanced by air resistance pushing upward. At terminal velocity, these forces are equal, so the skydiver stops accelerating and falls at a constant speed.