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.
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.
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).
The horizontal component of a projectile's velocity doesn't change, until the projectile hits somethingor falls to the ground.The vertical component of a projectile's velocity becomes [9.8 meters per second downward] greatereach second. At the maximum height of its trajectory, the projectile's velocity is zero. That's the pointwhere the velocity transitions from upward to downward.
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.
If, as you say, its acceleration is "constant", then the average is exactly equal to that constant.
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.
i will give u an illustration, consider an object projected (thrown)with some initial vertical velocity from the ground such that it traces a open downward parabolicpath, in that path the vertical displacement of the body from the point of projection to the point where it strikes the ground is equal to zero,but it have some velocity.
When an object is at terminal velocity, the two forces due to gravity and drag are equal, so the object ceases accelerating. Its motion is constant and vertically downward.
When an object is not accelerating or decelerating, it has a net force of zero.
It doesn't matter whether the object is thrown down, up, horizontally, or diagonally. Once it leaves the thrower's hand, it is accelerated downward by an amount equal to acceleration of gravity on the planet where this is all happening. On Earth, if you throw an object horizontally, it accelerates downward at the rate of 9.8 meters per second2 ... just as it would if you simply dropped it. Whether it's dropped or thrown horizontally, it hits the ground at the same time.
A the moment when the ball just touches the thrower's hand, it will have the velocity with which it was thrown and the acceleration will be equal to the acceleration due to gravity at the place acting vertically downwards.