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When a ball is thrown upwards and reaches its maximum height, does its weight get balanced by air resistances because it is stationary for a moment? Or does it have a force which is horizontal? In my book it is given as the net force acting in a horizontal direction.The question in my book is specific to the parabolic path of the projectile. But isn't the horizontal force acting on a projectile zero and isn't it why the horizontal component of the velocity is constant? Please correct me if I'm wrong. Thanks.

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  • $\begingroup$ Can you post the exact question given in ur book ? $\endgroup$ – 0xVikas May 11 '18 at 12:19
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When a ball is thrown upward and reaches maximum height, its weight is not balanced by anything - the gravitational force (weight) still acts, and accelerates it downward. What is special about the maximum height point is that the velocity vertical velocity is 0. This has nothing to do with air resistance. (which pushes the ball downwards before the maximum height point, with gravity)

You can have horizontal force due to air resistance if the ball has velocity in the horizontal direction, but it cannot cancel gravity, as gravity acts in the vertical direction.

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  • $\begingroup$ Thank you! What was bothering me is how the horizontal component of the velocity doesn't change when the air resistance does act on it? But I later realized that we always neglect the force of air resistance haha. $\endgroup$ – Anubhuti Srivastava May 11 '18 at 18:10
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There is a fundamental misunderstanding on force/motion here. For simplicity let's ignore air resistance, which is not needed for the ball to stop. Also, throwing the ball upwards is obviously a one-dimensional problem, so any "horizontal components" are of no interest here.

The physics of throwing the ball upwards is as follows: Just after you let go of the ball, the ball (of mass $m$) has an initial velocity, $v_0$, pointing upwards. There is only one force acting on the ball, which is gravity. The force is $F=mg$ (where $g\approx 10 m/s^2$ is the acceleration due to gravity) and it is pointing downwards, i.e. opposite to the initial velocity. This force is acting all the time and always has the same constant value (Actually the force gets slightly smaller the further you get from earth. However, unless you throw the ball very high, say into outer space, this is a very good approximation.)

According to Netwon's law this force acting on the ball will accelerate the ball with an acceleration $a=g$, which points into the same direction as the force, i.e. downward.

So what we have here is uniformly accelerated motion with initial velocity. I am sure you studied this one way or another. In any case you can write down the velocity for this motion as:

$v=v_0-gt$

The minus sign is here because the acceleration acts opposite to the initial velocity. Positive velocities are taken to be those pointing upwards and negative those that point down. As you can see from this equation, the velocity becomes zero after a time $t=v_0/g$ when the ball reaches the highest point. After that time, the velocity $v$ becomes negative, describing the ball falling down from the highest point.

So no magic here. In words, what is happening is that the force (gravity) is first decelerating the ball down to 0 and then accelerating it. For a physicist deceleration and acceleration are basically the same thing (only differ by the direction of acceleration relative to the velocity.


Regarding your other thoughts. Air resistance you only have on moving objects, so not if the velocity is zero.

For the parabola throw, the reasoning is that you can split the motion into two parts, a vertical component, which is identical to the vertical throwing discussed above, and a horizontal component which is just a motion with constant velocity, because there are no forces acting in the horizontal direction (ignoring air resistance...).

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