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March 5th, 2019, 01:19 PM  #1 
Member Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus  How to find the work in parabolic trajectory when the unknowns are angle and speed?
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 xaxis is always zero?. Why to bother finding it anyway? The ball's motion in xaxis doesn't change as it remains constant so there is no change in kinetic speed hence the work on $\textrm{xaxis}$ is zero. Isn't it? Is there any change of kinetic energy in the $\textrm{yaxis}$? There is change in kinetic energy in $\textrm{yaxis}$ 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: $710t=0$ $t=\frac{7}{10}$ (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). $y=7(\frac{7}{10})5\left(\frac{7}{10}\right)^{2}$ $y=\frac{49}{10}\frac{5\times 49}{10\times 10}$ $y=\frac{49}{10}\left(1\frac{1}{2}\right)$ $y=\frac{49}{10}\left(\frac{1}{2}\right)$ $y=\frac{49}{20}$ 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: $v^{2}_{f}=v^{2}_{o}2g(y)$ $v^{2}_{f}=0^{2}2(10)\left(\frac{49}{20}\right)$ $v^{2}_{f}=49$ $v_{f}=\sqrt{49}=7$ 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?. 
March 5th, 2019, 02:24 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,949 Thanks: 1555 
In a parabolic trajectory under the influence of gravity, the work done by the player is the initial kinetic energy given to the ball since it started from rest. The ball lands 14m from the spot it was kicked after 1.4 sec ... $\Delta x = v_x \cdot t \implies v_x = \dfrac{\Delta x}{t} = \dfrac{14 \, m}{1.4 \, sec} = 10 \, m/sec$ $v_x$, the horizontal component of velocity remains constant since only the force of gravity acting in the vertical direction is considered (drag forces due to air resistance are ignored) Since the ball completes its flight in 1.4 sec, it takes 0.7 sec to reach its highest point above the ground. At that point, velocity in the vertical direction is zero ... $v_{yf} = v_{y0}  gt \implies v_{y0} = v_{yf} + gt = 0 + (10 \, m/sec^2)(0.7 \, sec) = 7 \, m/s$ Initial velocity of the ball is $v_0 = \sqrt{v_x^2 + v_{y0}^2} = \sqrt{10^2 + 7^2} = \sqrt{149} \, m/sec$ $KE = \dfrac{1}{2}mv_0^2 = \dfrac{1}{2}(0.4 \, kg)(149 \, m^2/sec^2) = 29.8 \, kgm/sec^2 = 29.8 \, J$ 
March 6th, 2019, 05:37 PM  #3  
Member Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus  Quote:
First off, why isn't considered the $v_{o}\cos\omega$ in any of your equations for the xaxis and why $v_{o}\sin\omega$ is also left out from your analysis?. Does it makes sense to consider that the Kinetic energy solely on the $xaxis$ to be zero?. I mean since there isn't any change of the speed, wouldn't it mean that the $KE$ in this axis is zero?. Can be found the $KE$ on the $yaxis$ alone or should for this part the Energy must be calculated using the Gravitational potential energy $U$? You've found the $v_{oy}$ by using the equation in that axis, but why it doesn't consider the $\sin\omega$ function?. When we do analyze this situation isn't implied to use the trigonometrical functions for both axis?. I believe a graph or a sketch of the parabola with the right triangle would help me to understand why you took the square root to find the initial speed?. It makes sense from using the Pythagorean relationship but I want to know where are the directions of the vectors because this part I'm kind of confused, so can you please add some drawing?.  
March 6th, 2019, 06:36 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,949 Thanks: 1555 
Note ... $v_x = v_0 \cos{\theta_0}$ $v_{y0} = v_0 \sin{\theta_0}$ 

Tags 
angle, find, parabolic, speed, trajectory, unknowns, work 
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