# Forces on a ball thrown upwards

When a ball is thrown up in upward direction, it is said that force is in downward direction. Why we don't we consider the force given to the ball to throw up in the upward direction? Is there is no effect of the force given to the ball?

• During the throwing, the net force is upwards. Once you release it the only force is downwards. – Floris Mar 11 '15 at 17:26

You should have a look at Newtons First Law of Motion:

"When viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by an external force."

When the ball is moving up and there is no force at all, then the ball will continue it's motion upwards. But when there is gravity, you should look at Newtons Second Law of Motion:

"The vector sum of the forces $F$ on an object is equal to the mass $m$ of that object multiplied by the acceleration vector $a$ of the object: $F = ma$."

While the ball is moving up, you do not interact with it, therefore the only force is gravity, accelerating the ball downwards (slowing it down).

You should not confuse the concept of force with the concept of energy. When throwing the ball up, you accelerate it and therefore you transfer kinetic energy to the ball, not giving it force.

This is a classic misconception that most people share at some point in their lives. For centuries, we struggled to understand this point. For example, the famous Aristotle expresses your misconception that:

continuation of motion depends on continued action of a force

i.e. you see a ball moving upwards, and think that there must always be a force pushing it upwards. That is not the case. The ball has an initial velocity upwards, but the only force acting on the ball once it has left your hand is gravity.

Once the ball leaves your hand, it is moving upwards, but getting slower and slower, i.e. it is decelerating (or accelerating downwards). This deceleration is caused by gravity, a force acting downwards.

Nowadays, this fact is trivial, known by millions, but it was a significant development in the history of physics that confused some of the most famous minds.

• now worries @shyam, hope it's clear now – innisfree Mar 12 '15 at 9:11
• YES, thanks a lot for helping me to clearify my long time confusion @innisfree – SHYAMANANDA NINGOMBAM Mar 12 '15 at 16:20

Once you release the ball, you are not applying a force to it; it is freely falling (despite its upward motion). The only force acting on it is the gravitational force, pulling it downwards (which is why it slows down and stops momentarily at the apex, before coming back down).

See also the related question When a ball is tossed straight up, does it experience momentary equilibrium at top of its path?

• then if it is freely fall,act only by gravity why it moves upward? – SHYAMANANDA NINGOMBAM Mar 11 '15 at 17:08
• Because the force you applied to the ball to throw it was greater than $mg$ (otherwise it could not move); it then continues decelerating as it moves upwards. – Kyle Kanos Mar 11 '15 at 17:16
• @SHYAMANANDA NINGOMBAM :By freely falling he means the body moves only under the presence of gravity(not any other forces)and the body is moving upwards because the you initially provided a kinetic energy. – Paul Mar 11 '15 at 17:18

Yes,the effect of force with which the ball is thrown is surely considered. When the ball is thrown upwards,the gravity try to pull it downwards and hence after sometime it starts coming down(when gravitational force overcome the effect of force you provided when you threw it).Somehow,for ease of solving,at lower level it is assumed that the only force acting on ball after being thrown is gravitational force.

Actually, when you throw the ball up into the air the ball will not only be acted on by gravity but friction and resistance from the air molecules colliding with the movement of the ball.

Inao Shyamananda. The act of throwing a ball upward can be studied in two stages.

Stage 1: When you throw a ball up you apply a force to the ball in the upward direction as long as it is in contact with your hand. This force does some amount of work on the ball. This work done is manifest as the sum of kinetic energy and potential energy of the ball. At the end of this stage the ball leaves your hand. The kinetic energy at the end of this stage gives you the "initial velocity" for the next stage through the relation $$KE=\frac{1}{2}mv^2.$$ I hope this clears your doubts about what happened to the force which your hand exerted on the ball.

Stage 2: The moment the ball leaves your hand, the only force acting on the ball now is the gravitational force (neglecting the effect of air). This stage has been clearly explained in the earlier answers.