Notice Since I have written the majority of this answer the question changed a lot, and now it's been put on hold. However, I won't erase parts of my answer which are answers for the original (ill posed) question, because other people might find it useful. I have extended it with parts though, which seem (to me) to answer the question the original changed to.
I like David Hammen's and a ACuriousMind's answers, but since this question scratches the philosophy behind classical mechanics, I can't resist writing my own line of thoughts. I will be gradually writing and updating this answer however, because I'm in the middle of exams, and don't have much time.
So, to begin with, as David Hammen pointed out, one can't ignore Newton's other three axioms and just select the second. More to come here, but for now, as a reminder, please note that Newtons laws only hold for pointlike bodies. If the spatial extent of a body cannot be neglected, it is to be treated as infinitely many, infinitely small points, and one has to treat it with methods of calculus. Now let's just jump to the second axiom.
Newton's second law of motion is actually incorrect in the form $F=ma$. First of all, it's a vectorial equation, so it should be $\vec{\mathbf{F}}=m\vec{\mathbf{a}}$. This takes care of the misunderstanding that direction isn't specified. It inherently is. The law is still incorrect in this form however, and even though Newton did not about special relativity, he wrote it in the form (to be precise what he wrote is translated algebraically as) $\vec{\mathbf{F}}=\frac{d\vec{\mathbf{p}}}{dt}$ where $\vec{\mathbf{p}}$ is the momentum, defined to be $\vec{\mathbf{p}}=m\vec{\mathbf{v}}$. This is very important since special relativity tells you that if a body has mass $m_0$ in a reference frame in which it is at rest, it has a mass $m=m_0/\sqrt{1-v^2/c^2}$ in a reference frame from which it is seen moving with speed $v=\left|\vec{\mathbf{v}}\right|$. But since in Newton's time any achievable speed was well below lightspeed, $v^2/c^2\approx0$ thus $m(v)\approx m_0$. Hence the second law reads as \begin{equation}\vec{\mathbf{F}}=m\frac{d\vec{\mathbf{v}}}{dt}=m\vec{\mathbf{a}}=m\frac{d^2\vec{\mathbf{r}}}{dt^2}=m\ddot{\vec{\mathbf{r}}}.\end{equation}
One should immediately ask him/herself (as you did) what this law really tells one. It is a second order differential equation that actually connets an action of the surrounding environment ($\vec{\mathbf{F}}$) with the change in motion of a body ($\ddot{\vec{\mathbf{r}}}$). It also tells you that whatever the action of the sorrounding is, the change in motion is connected with it linerly, and the proportionality factor is a parameter characteristic of the pointlike body ($m$).
Firstly, this very law combined with the third law (that is a force exerted by a point on another is always attached with a reaction force equal in magnitude, opposite in direction acting on the former point, exerted by the latter point) lets you measure the proportionality factor $m$. Eg. you take a spring (doesn't matter how strong) and a billiard ball which you specify as having mass $m=1~unit$ and you put them on a table treacled with oil. Then put the ball on one end of the spring, put the object with mass to be measured on the other, and let the spring exert force on each of them. This will be the same force because of the third law (you don't even have to know how the spring behaves). By measuring how far each of them went, you can measure how many units of mass the other body had (the friction that remains despite the oil gives the error in the measurement).
Now more deeper into your question.
But the most striking feature of the formula F=ma is that it doesn't say anything about the applied force, but can only measure its oucome.
In this, you are perfectly right. $\vec{\mathbf{F}}=m\vec{\mathbf{a}}$ is completely useless until you don't know what to write in the l.h.s. Newton's ingeniality lied in not only giving the relationship between the change in the state of motin and the action of the environment, but also in giving formulas about how the action looks like in specific cases (e.g. for a spring it's $\vec{\mathbf{F}}=-k\vec{\mathbf{\Delta r}}$).
This takes us to the notion of "law of force" (this is raw translation from my mother language, please write me its proper English name, and I will correct it). These are the laws that tell you what to write in place of $\vec{\mathbf{F}}$ in the case of specific interactions. E.g. in the case of a spring it is $\vec{\mathbf{F}}=-k\vec{\mathbf{\Delta r}}$. In case of gravitational interaction it is $\vec{\mathbf{F}}=-G\frac{M_1M_2}{r^2}\frac{\vec{\mathbf{r}}}{r}$ (this was another ingenious guess by Newton). Of course many of these laws are incorrect in the sense that they don't describe the full behavior of an object. E.g. you pull a string of $1~cm$ in length to a meter, and it will surely not have a linear behaviour. Pull it to a light year in lenght and it will surely be torn apart. These laws of force are mostly the results of measurements, and in many cases you cannot derive them from first principles, or you have to make guesses about the behavior of the interaction you study. E.g. in an elastic medium you may suppose that if you pull or twist it, the forces acting between two points are like there were springs between them. If you wanted to be more precise, you would have to use quantum mechanics, and directly derive the interactions between atoms, but still there would be parameters, even potentials in your equations which have been generalized from results of measurements. In physics, in the end, there's no way around using some results, even laws, acquired from measurements.
As ACuriousMind has implicitly pointed out, even until today, we haven't found any force laws that can not be written in the form \begin{equation}\vec{\mathbf{F}}=\vec{\mathbf{F}}\left(t, \vec{\mathbf{r}}(t), \dot{\vec{\mathbf{r}}}(t)\right).\end{equation}
This means that the equation of motion (as we know, this is also a law of nature) can not be a differenctial equation with higher order than second. You didn't ask this, but this is also important, that specifying this differenctial equation (DE), ie. the equation of motion, is not enough to give a correct description of the motion of the object, since it's solution alway contains two free parameters (a result from the theory of DE's). These are specified by two restraining conditions, which you also have to provide by measurement or by some reasoning outside Newton's axioms. E.g. they may be intial conditions, such that the object which you study was at time $t=0$ in $\vec{\mathbf{r}}(0)=\vec{\mathbf{r}}_0$ and it had velocity $\vec{\mathbf{v}}(0)=\vec{\mathbf{v}}_0$. Two independent restraining conditions and the equation of motion are always enough to sepcify the trajectory of a body under study.
As you can see the question is very deep, the answer spans several pages, but you asked for it, so just keep on reading.
- we know nothing of the energy that was spent,
- of the energy the object will acquire;
- we know nothing of the change of velocity of the object, and
- we can't even be sure that the change of momentum will actually be 50, since there is a limit to the acceleration due to the max v reached by the hand/ fingertips (therefore the momentum of a ping-pong ball can be at most, say, 1 kgm/s).
Now this is absolutely not true. The basic laws are Newton's axioms, energy is derived from them (some people derive momentum from them, but I prefer taking momentum as a fundamental quantity, as Newton did). I will show these through the example you asked for.
- F⃗ =120N for a certain time
- t=.25sec on a body of mass (M)
- MB=0.1Kg
What is the change in velocity of the body? MB=0.1
Please note that by giving this exercise you implicitly (but I think not consciously) supposed the following two simplifying guesses and conditions: you don't care about the direction of the force, which makes the calculation more easy. By not caring about direction, you also suppose that there is no restraint on the trajectory of the thrown ball, thus we don't have to care about things like just some component of the pushing force matters. E.g. if you wanted to push a train, or push a ball into a pipe, only the force component parallel with the ground and rails or in case of the ball, tube would matter. Of course this is also a simplification, since if you applied great enough force component in other directions, the pipe would break, or you would tip over the train. Secondly you supposed that the force is constant (independent of time AND place AND velocity), which I will put into use immediately. Thirdly, and most importantly, you supposed drag is neglectable, which is not, as you will see immediately.
First, the momentum. As you have seen $\vec{\mathbf{F}}=d\vec{\mathbf{p}}/dt$. which is easily inversed by integration: \begin{equation}\int_{t_0}^{t_0+\Delta t}\vec{\mathbf{F}}\left(t, \vec{\mathbf{r}}(t), \dot{\vec{\mathbf{r}}}(t)\right)dt = \vec{\mathbf{p}}(t_0+\Delta t)-\vec{\mathbf{p}}(t_0)=\Delta\vec{\mathbf{p}}\end{equation} Supposing that the force is constant: \begin{equation}\int_{t_0}^{t_0+\Delta t}\vec{\mathbf{F}}\left(t, \vec{\mathbf{r}}(t), \dot{\vec{\mathbf{r}}}(t)\right)dt = \int_{t_0}^{t_0+\Delta t}\vec{\mathbf{F}}dt = \vec{\mathbf{F}}\Delta t\end{equation}
Thus $\vec{\mathbf{F}}\Delta t = \Delta\vec{\mathbf{p}}$. We don't care about directions and suppose everything is fine, so this can be considered as a scalar equation: $F\Delta t = \Delta p$. Given you example the change in momentum and thus speed (since mass is supposed not to change) will precisely be \begin{equation}120*0.25=0.1*\Delta v\end{equation} \begin{equation}\Delta v=300~m/s\end{equation} acquired from $F=ma$.
As you have pointed out in the comments (which you also have erased since then) this is an unreasonable fast ball. Let's do the calculation again without neglecting drag. The drag for relatively high Reynolds numbers ($R \geq \approx 1000$) is expressed as \begin{equation}F=\frac{1}{2}\rho v^2 C_D A,\end{equation} where $\rho$ is the density of air ($\approx 1.225~kg/m^3$), $v^2$ is the realtive speed of the object moving through it, $C_D$ is the drag coefficient ($0.47$ in case of a spherical object) and $A$ is the cross sectional area. For low Reynolds numbers (viscous fluid, slow speed, which is clearly not the case for a thrown ball), the drag is described by \begin{equation}F=-6\pi\eta r v.\end{equation} Reynolds number is written as \begin{equation}\frac{\rho}{\eta}vl,\end{equation} where $v$ is again the speed of the object, $\rho$ is air density, $\eta$ is air viscosity ($\approx 0.7978*10^{-3}~Pa\cdot s$) and $l$ is a characteristic size of the object. $R$ for a basketball with $\approx0.2~m$ radius ($\to~l\approx2*0.2$) is $R=\frac{1.225}{0.7978*10^{-3}}*0.4*v\approx v*614~s/m$. It is clearly visible that the ball basically immediately reaches $R~>1000$, so it is okay to use the formula for drag with $v^2$ in it. This gives a new equation of motion (taking the mass of the ball to be $m=0.1~kg$): \begin{equation}m\ddot{x}=120-\frac{1}{2}\rho v^2 C_D A\end{equation}\begin{equation}\ddot{x}=1200-1.4v^2\end{equation}\begin{equation}\dot{v}=1200-1.4v^2\end{equation} Solved by WolframAlpha with initial condition $v(0)=0~m/s$, the result is: \begin{equation}v(t)=\frac{29.277e^{81.9756t}-29.277}{e^{81.9756t}+1}\end{equation} This equation of motion was valid until $0.25~s$, because after that, the $120~N$ push became zero. Substituting $0.25~s$ to get the velocity the ball reaches after leaving the thrower's hand we get \begin{equation}v\approx30~m/s,\end{equation} which is quite reasonable.
Please note, that it is right that there is a limit on the acceleration human fingers are capable of, but it was YOU who told me that the force is $120~N$ for a time $t=0.25~s$. If you wanted a more precise theory of throwing a ball, you should study biomechanics of human muscles, and than acquire reasonable force and time data from there. Basically you would have to invent the force law of a human throwing a ball, which as you have noticed, depends on some biomechanical parameters such as the max. velocity achievable by someones fingers, etc.
Energy comes now. Energy is obtained by \begin{equation} \vec{\mathbf{F}} = m\ddot{\vec{\mathbf{r}}}~/\text{multiply both sides by } \vec{\mathbf{v}} \end{equation}
\begin{equation}\vec{\mathbf{F}}\vec{\mathbf{v}} = m\ddot{\vec{\mathbf{r}}}\vec{\mathbf{v}}=\dot{\vec{\mathbf{v}}}\vec{\mathbf{v}}=\frac{1}{2}m\frac{d\vec{\mathbf{v}}^2}{dt} \end{equation}\begin{equation}P = \frac{1}{2}m\frac{d\vec{\mathbf{v}}^2}{dt}\end{equation}\begin{equation}W=\int_{t_0}^{t_0 + \Delta t}Pdt = \int_{t_0}^{t_0 + \Delta t}\vec{\mathbf{F}}\vec{\mathbf{v}}dt = \frac{1}{2}m\left[\vec{\mathbf{v}}^2(t_0+\Delta t)-\vec{\mathbf{v}}^2(t_0)\right]=\Delta E_{kin}\end{equation} The change in in kinetic energy is obtained by either integrating the force multiplied by velocity $(120-0.14*v^2)*v$ form $t_0=0~s$ to $t_0+\Delta t=.25~s$, or simply by squaring the final velocity and multiplying it by $m/2$: \begin{equation}\Delta E_{kin}\approx45~J,\end{equation} but the energy spent was much more, because most of it was spent to counter friction. The total energy (work) spent is obtained by integrating only the $120~N$ part multiplied by $v(t)$: \begin{equation}W_{sum}\approx818.897~J\end{equation} Again, the energy burnt by muscles is even more, because energy conversion from glucose's chemical energy to kinetic energy is far from perfect.
I think this rather well cleans up the mess around why doesn't a ball reach enormous acceleration due to the hughe force a thrower exerts.
