Neither the acceleration nor the velocity of the ball will be zero during its flight.
The motion of a kicked soccer ball is actually rather complex, and we'll have to simplify the system a bit and make some assumptions if we're going to be able to talk about it here without using advanced mathematics or complex computer simulations.
So let's get those simplifications and assumptions out of the way, shall we?
We will ignore the effects of the air on the ball. In other words, we're assuming the ball is traveling through a vacuum. This is important for two reasons. One is we can assume the ball's horizontal velocity doesn't change from the moment it leaves the player's foot to the moment it lands in the back of the net or the goal tender's hands. This is because we're ignoring the effects of air resistance.
The second is we can forget about the cool "bending" tricks that great players (like Beckham) can make the ball do by putting spin on it. So, for our discussion, the ball moves in two dimensions: the vertical (the y direction) and the horizontal (the x direction). We can ignore the third direction: the z direction. Hence, the ball follows a perfect parabolic arc in one plane.
When the ball leaves the player's foot, it has an initial velocity that has both vertical and horizontal components. The horizontal component propels the ball down the field; the vertical component propels the ball up into the air. Let's look at the vertical component, first.
From the split second the ball leaves the player's foot, the ball experiences an acceleration in the downward direction. Yes, downward! Even though the ball is heading up, the acceleration it experiences is down! That is because the acceleration due to gravity is always pointed toward the center of the Earth -- in other words, DOWN! This causes the ball to slow down, but only in the vertical direction. The vertical velocity decreases until the ball reaches its apex, at which point the vertical velocity is zero. After it reaches its apex, the ball starts to accelerate downward. In other words, it picks up speed in the downward direction.
How about the horizontal component? Well, since we're ignoring air resistance, it doesn't change for the whole flight. No kidding. If the ball's horizontal velocity is 30 meters per second as it leaves the player's foot, it will be 30 meters per second for the whole flight. So, you can see that the vector sum of the vertical and horizontal components of the velocity is never zero, since there is always a non-zero horizontal component.
The discussion of acceleration is trivial. The acceleration of gravity is always there. It never goes away -- let's HOPE it never goes away!! -- so the ball always experiences it. And it's always directed downward, even when the ball's going up.
To summarize, the acceleration and velocity of the ball is always non-zero.
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1. This is somewhat counter-intuitive and hard for new physics students to get their brains around. The acceleration of gravity is perpendicular (orthogonal) to the horizontal motion. Therefore, there is no horizontal component of the acceleration due to gravity. Because there is no horizontal component, it cannot affect the horizontal motion of the ball.
2. The spin on a ball in a medium like air creates pressure differentials that produce forces on the ball that make it do funny things. The same principle is at work when a Baseball pitcher throws a curve ball, slider, or split-fingered fastball: the spin makes the ball move in the direction of rotation.
3. The ball's vertical velocity is zero only for an instant, an instant so brief as to be too small to quantify. The velocity at any instant that is too small to quantify is called the instantaneous velocity. (That should be easy to remember.)
4. By now you may wondering why I keep saying "as it leaves the player's foot" or "a split second after the ball leaves the player's foot." Well, that is because we're only talking about the motion of the ball after the player has finished doing work on the ball -- that is, KICKING it! It's a whole other physics problem to analyze what happens to the ball during the act of kicking it. The force applied to the ball during the kick is what accelerates it and gives it its initial velocity for the problem being discussed here. That acceleration has NOTHING to do with the acceleration of gravity.
Yes, at the highest point of the projectile's trajectory, the velocity and acceleration vectors are parallel to each other. This is because the velocity is momentarily zero, and the acceleration due to gravity is acting vertically downward, pointing in the same direction as the velocity.
For example, an object thrown upwards, when it is at its highest point. This situation is only possible for an instant - if the acceleration is non-zero, the velocity changes, and can therefore not remain at zero.
The vertical velocity at the highest point of the trajectory, the vertical displacement when the projectile returns to its initial height, and the vertical acceleration at the highest point are all zero throughout the flight of a projectile.
The acceleration of the ball (after it leaves the thrower's hand) is the acceleration due to gravity, g.1 The vertical velocity of the ball at its apex is zero. The horizontal velocity is constant throughout the ball's flight; it is whatever it was at the outset of its arc.2 ---------------- 1. The acceleration due to gravity, g, is -9.8 m/s2 or -32.2 ft/s2. 2. Ignoring the effects of air resistance, which tend to slow things down.
At the highest point in its trajectory, the vertical velocity of a projectile is zero. This is because the projectile has reached its peak height and is momentarily at rest before starting to descend.
Using the term "trajectory" implies that the acceleration you are concerned about is due to gravity. Gravity will always be perpendicular to the surface. Unless the trajectory begins perpendicular to the surface, it will never change to become perpendicular and the velocity will never be in a direction parallel to the acceleration. If it starts perpendicular to the surface it will start and remain perpendicular. Of course if you have another force acting on the object - such as wind - the component of the velocity vector parallel to the ground could be reduced to zero and at that point the only remaining component of the velocity vector would be that perpendicular to the ground and parallel to the acceleration. Likewise if the object is being propelled by an engine or rocket, the trajectory could be parallel to the force any time the acceleration vector became parallel to the velocity vector.
Answer:Yes, but only instantaneously.Consider a thrown ball moving directly upward. At the highest point of its trajectory, the instanataneous velocity (the velocity at that precise instant) is zero even while the acceleration due to gravity remains non zero.
At the highest point of its trajectory, the direction of an oblique projectile will be horizontal. This means that the projectile will momentarily have zero vertical velocity and only horizontal velocity.
The vertical speed at the highest point of a projectile's trajectory is zero. This is because at the peak of the trajectory, the projectile momentarily stops ascending and starts descending, resulting in a velocity of zero in the vertical direction.
Yes, but only for an instant. For example, if you throw a stone up, when it is at its highest point it has a velocity of zero, but its acceleration is -9.8 m/s2. If there is acceleration, the velocity can not remain at zero.
Just before it reaches the highest point, the vertical component of velocity is upward.Just after it passes the highest point, the vertical component of velocity is downward.There's no way you can change from an upward velocity to a downward velocity smoothlywithout velocity being zero at some instant. A.True.
The acceleration is the acceleration of gravity, downwards, or 9.8m/s/s (32 ft/s/s). When ball is thrown straight up it has an initial velocity that is decreasing because of gravity; at the highest point velocity is zero but acceleration is always constant at gravity rate.