Do we need to exert a force larger than the weight of an object in order to raise it a certain height? I had a problem in my book stating:

A boy lifts a ball with mass $m$ with force of constant magnitude $F$ to a height $h$.
Calculate the force magnitude $F$ acting on the ball .

I know this is a very silly question. But I am screwing up in understanding the answer in my book. Please help me!
Here,the answer given in my book is $mg$. But I think the force must be greater than that.
To explain my answer, I have taken the scenario as follows: If the boy is holding the ball in his hand, then the force of gravity and the force by his hand counter balance each other as a result of which ball stays stationary. Now, if the boy has to move the ball to a greater height, he must apply an additional force along with the force that keeps the ball stationary. Thus, my answer.
So, can the force being equal to the weight raise the ball up to a height $h$, or does the force need to be larger than this?
 A: Yes, a very poorly worded question. First of all, we must make some assumptions. So let’s assume

*

*the force $F$ is applied directly to the ball

*the ball is lifted vertically through a height $h$

*the ball is stationary at the start and the end of its motion

*there are no friction or drag forces

Note that none of this is stated explicitly in the question.
With these assumptions then, as you say, to get the ball moving in the first place the initial force applied to the ball must be greater than $mg$. And if we want the ball to be stationary again at height $h$ then at some point in the motion we need to reduce the force to less than $mg$ so that the ball slows down again. In this case all we can say, even with these assumptions, is that the average force applied to ball as it is raised is $mg$.
And, of course, if the ball does not need to be stationary at height $h$ then we can raise it with any direct force we like that is greater than $mg$.
Further more, if the force is not applied directly to the ball or the ball is not lifted vertically, then the boy may raise the ball with a force less than $mg$ if he applies it over a greater distance, using, for example, a lever or an inclined plane.
So the main educational benefit of this question seems to lie in thinking of all the ways in which the given answer can be wrong !
A: I think the key here is that the question states that the force is 'of constant magnitude', and yet it moves the ball.
Whenever you accelerate an object away from you you increase the force with which you press on it, and whenever you let it decelerate you decrease the force. As the force is stated to be of constant magnitude we must assume that the acceleration is so small that we can ignore it when calculating the force. If we did not ignore acceleration the force would be higher at the start as the ball is accelerated upwards and lower at the end as it is decelerated.
Or alternatively maybe the acceleration is so brief that we can ignore it.
Once we've ignored acceleration, the force the boy applies to the ball must exactly match the force the earth applies to it as gravity in the opposite direction, so the magnitude is mg.
A: This is a horribly written question (the exercise, not your post).
You are correct in your reasoning. If the ball starts at rest, the force $F$ needs to be larger than $mg$ in order for it to begin moving upwards.
However, if the ball already starts with some initial upwards velocity, a force $F=mg$ would be enough to get the ball to keep moving upwards, and any force larger than $mg$ still gets the job done. With a sufficiently large initial velocity, the force $F$ could actually be smaller than $mg$, since all we require is the ball move to a height $h$ rather than move to a height $h$ and then still keep moving upwards.
So, to summarize, $F>mg$ always works, and $0<F\leq mg$ works for a sufficiently large initial upward velocity$^*$.
The issue with the problem is that it doesn't specify the initial conditions of the ball, and even if they had been specified there is not a unique answer; there is just a minimal force needed to raise the ball a height $h$. So yeah, ignore the fact that you didn't get the answer and instead focus on the concepts that you have correct. $F=mg$ means no acceleration, and if the ball starts at rest this means, in this case, it could not reach height $h$ with only $F$ and the weight acting on it.

$^*$ F could technically be negative with a sufficiently large initial upward velocity, but since we are told the ball is being lifted we will assume the force has to be upwards.
A: The questioner stated that "But I think the force must be greater than that." which is of course the correct answer.
The basic equation which defines no movement of the ball would be F=mg.   Without complicating the question any vertical upward component of applied force greater than mg will move the ball upwards. If this vertical component of applied force is a constant the ball will keep accelerating upwards as long as F>mg and so this simple equation F>mg will satisfy all of the terms included in the question.
A: As the time interval in which the force is applied is also important, I will reword the question as follows.

An object of mass $m$ moving vertically upward and decelerating due to gravity $g$ has an initial velocity $u\ge 0$  at $t=0$. If the object must reach a point $h>0$ above it at $0< t \le  T$, find the minimum additional upward force $F$ needed.

Case 1: $u=0$
\begin{align}
F-mg &=ma\\
h&=\frac{1}{2}aT^2
\end{align}
The minimum force is $F=m\left(\frac{2h}{T^2}+g\right)$.
Case 2: $u>0$ and $T\ge\frac{u}{g}$ and $h\le\frac{u^2}{2g}$
The minimum force is $F=0$. Let the object continue moving and it is guaranteed that it will reach the target point.
Case 3: Otherwise
\begin{align}
F-mg &=ma\\
h&=uT + \frac{1}{2}aT^2
\end{align}
The minimum force is $F=m\left(\frac{2(h-uT)}{T^2}+g\right)$.
