I'm confused about this question from my book. Apparently it is supposed to be simple but upon reading it several times I don't know how to proceed. I need also assistance with some theoretical or conceptual knowledge on how to find the work in a parabolic trajectory.
The problem is as follows:
A soccer player kicks a ball from rest over the head of a rival. The ball falls $\textrm{14 m}$ from the player who kicked the ball after $\textrm{1.4 s}$. At a TV studio the sports commentator decides to calculate the work done by the player in that pass. The sports programme tells the audience that the official ball has a mass of $\textrm{400 grams}$ and the trajectory is a parabola. Find the work calculated by the sports show commentator. Consider the gravity is $10\frac{m}{s^{2}}$
$\begin{array}{ll} 1.&\textrm{27.25 J}\\ 2.&\textrm{32.8 J}\\ 3.&\textrm{35 J}\\ 4.&\textrm{17.5 J}\\ 5.&\textrm{29.8 J}\\ \end{array}$
From the knowledge I have. In order to find the work done in this case it would be from the fact that this is given by the difference in the kinetic energy of the ball.
The reason for this part of my judgment is due that when the ball touches ground there is no potential energy but there is kinetic energy.
There is the fact that the energy of the soccer player which is needed to find the work, is the same that the energy he uses to kick the ball and that's the energy the ball has.
Hence it would meant that what is needed to be found is the difference in kinetic energy the ball has so the work can be found.
In my attempt to find the launching angle I tried this:
$$y = y_{o}+v_{o}\sin\omega-\frac{1}{2}gt^2$$
Since it is given $t=1.4$ and $g=10$, (for brevity purposes I'm omitting units)
$$0=v_{o}(1.4)\sin\omega-\frac{1}{2}(10)(1.4)^2$$
$$v_{o}\sin\omega=(5)(1.4)$$
The other known is given, when $x=14$:
$$x=v_{o}t\cos\omega$$
$$14=v_{o}(1.4)\cos\omega$$
Thus:
$$v_{o}\cos\omega=10$$
From this can be obtained by dividing both equations:
$$\frac{5(1.4)}{10}=\tan\omega$$
Therefore:
$$\tan\omega=\frac{7}{10}$$
From this I believe the initial speed can be calculated from knowing the $\cos\omega$ which can be obtained from the previous equation as follows:
$$\cos\omega=\frac{10}{\sqrt{149}}$$
Therefore:
$$v_{o}=\frac{10}{\frac{10}{\sqrt{149}}}=\sqrt{149}$$
Now here's where I'm stuck at:
How do I make the right interpretation for finding the Work?
Would I go on this route?
$$W=\frac{1}{2}mv^2=\frac{1}{2}(400)(10^{-3})(\sqrt{149})^2=(2)(10^{-1})(149) = \textrm{29.8 J}$$
So the work found by the sportscaster would be $29.8\,J$
This answer does seem to check with one of the alternatives which is the fifth and my book tells this is the right answer, but I'm not very convinced if that would be the right one. Needless to say If what I'm doing is correct.
Looking at my intuition I have these other questions.
Does the kinetic energy in the x-axis is always zero?. Why to bother finding it anyway?
The ball's motion in x-axis doesn't change as it remains constant so there is no change in kinetic speed hence the work on $\textrm{x-axis}$ is zero. Isn't it?
Is there any change of kinetic energy in the $\textrm{y-axis}$?
There is change in kinetic energy in $\textrm{y-axis}$ I presume. But If the ball goes up in the apex of the trajectory it will have zero in its speed. So, when the ball touches ground will it have the same speed when it departed?. Can this be proved?
I hope somebody can help me with these doubts and more importantly offer some other alternative to easily solve this problem.
By continuing in this problem I think the height can be found knowing the $v_{o}=\sqrt{149}$ and $\sin\omega=\frac{7}{\sqrt{149}}$.
$$y = y_{o}+v_{o}t\sin\omega-\frac{1}{2}gt^2$$
Assuming kicking from ground so $y_{o}=0$
$$h= 7t - 5t^2$$
Using derivatives:
$$7-10t=0$$
so $t=\frac{7}{10}$ (although it was known from the problem that since it took $1.4$ seconds to get the whole trajectory then the apex would had been half of that time hence $0.7$ seconds).
$$y=7(\frac{7}{10})-5\left(\frac{7}{10}\right)^{2}=\frac{49}{10}-\frac{5\times 49}{10\times 10}=\frac{49}{10}\left(1-\frac{1}{2}\right)=\frac{49}{10}\left(\frac{1}{2}\right)=\frac{49}{20}$$
Therefore the height attained by the ball would be $\frac{49}{20}$ meters.
If I decide to use this to find the speed which will have the ball when reaches bottom I would use:
This would be:
$$v^{2}_{f}=v^{2}_{o}-2g(y)=0^{2}-2(10)\left(-\frac{49}{20}\right)=49$$
and so $$v_{f}=\sqrt{49}=7$$
Which it would meant 7 meters per second and this doesn't seem any close with what I assumed, in other words if the ball is kicked at some speed would it touch ground at that same speed.
However it checks if I would use:
$v_{f}=v_{o}-gt=0-(10)(0.7)=-7$
So in absolute value both are the same.
Overall can somebody explain to me what's happening here? and clear out the doubts which I mentioned above?
Obviously your calculations seem right, since you get the correct answer. So what you are missing is some intuition about why are you doing the steps you wrote about.
You might know that gravity is a conservative force. What does it mean? It means that no matter what trajectory you move an object, if it's at the same height it implies that the potential energy is the same. Now, for an object that moves only due to gravity, if it starts and end at the same height it means that the kinetic energy is the same (the magnitude of the velocity is the same).
Here are the replies to your questions:
You did not ask, but you might have wondered, where is the work being done if the potential energy is the same, and the kinetic energy is the same? The answer is that it happened when the ball was kicked. It says it started from rest, and immediately acquired a velocity $v_0$. So the change in kinetic energy, $\frac 12 mv_0^2$, all happened in a very short time period, at the contact between the ball and the foot.