April 13th, 2015, 05:30 AM  #1 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty  Newton's Laws: Question 2
An exceptional standing jump would raise a person $\displaystyle 0.80m$ off the ground. To do this, what force must a $\displaystyle 68kg$ person exert against the ground? Assume the person crouches a distance of $\displaystyle 0.20m$ prior to jumping and thus the upward force has this distance to act over before he leaves the ground. 1) F exerted over 0.20m $\displaystyle F = m\cdot a$ $\displaystyle F = 68\cdot 9.8$ $\displaystyle F = \frac{3332}{5}N = 666.4N$ 2) F exerted over 0.80m $\displaystyle v^2 = v_0^2 + 2a(yy_0)$ $\displaystyle v = \sqrt{0^2 + [2\cdot 9.8\cdot 0.20]}$ $\displaystyle v = \sqrt{\frac{98}{25}}m/s = 1.98m/s$ $\displaystyle v^2 = v_0^2 + 2a(yy_0)$ $\displaystyle a = \frac{\bigtriangleup v^2}{2\bigtriangleup y}$ $\displaystyle a = \frac{\frac{98}{25}}{2\cdot 0.80}$ $\displaystyle a = \frac{49}{20}m/s^2 = 2.45m/s^2 = a_R = a_{up} + a_{down} = a_{up}  g$ we want to know $\displaystyle a_{up}$ therefore $\displaystyle a = \frac{49}{20} + g$, $\displaystyle a = 12\frac{1}{4}m/s^2$ 3) Total force exerted $\displaystyle F = 68\cdot 12\frac{1}{4}$ $\displaystyle F = 833N$ My textbook gives an answer of $\displaystyle 3.3\times 10^3N $ which coincidentally is $\displaystyle 4 \times 833$. Where have I gone wrong? My reasoning for my solution is: 1) By crouching, he lowers his centre of mass $\displaystyle 0.20m$. Upon extension, his centre of mass accelerates $\displaystyle 9.8m/s^2$ upward, over a distance of $\displaystyle 0.20m$. I chose $\displaystyle a = g$ in this instance because his feet have not yet left the ground. I think this is where my mistake is. 2) He launches with an initial of $\displaystyle v_0 = 1.98m/s.$ Had he not crouched, his launch velocity $\displaystyle v_0 = 0m/s$. At 0.80m his velocity $\displaystyle v = 0m/s$. 3) His acceleration over $\displaystyle 0.80m$ is $\displaystyle a = 2.45m/s^2 $however this is $\displaystyle a_R$ ($\displaystyle a_R = a_{up}  g$) so $\displaystyle a = 12\frac{1}{4}m/s^2$. Overall, the person must over come gravity in order to jump $\displaystyle 0.80m$ off the ground therefore, $\displaystyle F_{exerted} > F_{gravity}$. My solution agrees with this but is still incorrect. Last edited by hyperbola; April 13th, 2015 at 05:35 AM. 
April 13th, 2015, 06:06 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,161 Thanks: 734 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Try using conservation of energy instead of SUVAT... The gravitational potential energy required in getting someone's feet to a height of 0.8m from 0m is $\displaystyle W = mg\Delta h = 68 \times 9.81 \times 0.8 = 533.664 J$ The gravitational potential energy required in getting someone's centre of mass from a crouching position to a standing position is $\displaystyle W = mg\Delta h = 68 \times 9.81 \times 0.2 = 133.416 J$ Total work done $\displaystyle = 533.664 + 133.416 = 667.08 J$ The person jumping must exert a force such that the amount of work done is equal to this (converting chemical energy to kinetic energy to gravitational potential energy) $\displaystyle W = Fd$ $\displaystyle F = \frac{W}{d} = \frac{667.08}{0.2} = 3335.4 N$ Note that this solution is assuming that someone goes over the bar in a standing posture, which would make for an odd looking jump! 
April 13th, 2015, 06:13 AM  #3 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty 
Thanks again Benit13. I haven't reached the energy section of my textbook yet. I'm still in the Newton's Laws section so therefore am working off the formulas given in this particular section and previous sections (Kinematics). However $\displaystyle mg\bigtriangleup{h}$ and $\displaystyle W=F\cdot{d}$ does ring a bell from high school. Thanks. Last edited by hyperbola; April 13th, 2015 at 06:18 AM. 
April 13th, 2015, 06:45 AM  #4 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty 
Benit13, Given your solution $\displaystyle F = 3335.4N$. The person accelerates, $\displaystyle a = \frac{F}{m} = \frac{3335.4}{68} = \frac{981}{20}m/s^2 = 49.05m/s^2$ That's superhuman. 
April 13th, 2015, 07:13 AM  #5 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,161 Thanks: 734 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
It is an exceptional standing jump after all! Notice that 5gs is very large, but not superhuman. A professional cricketer batting a cricket ball can achieve accelerations of 10,000 m/s$\displaystyle ^2$ and more! Last edited by Benit13; April 13th, 2015 at 07:22 AM. 
April 13th, 2015, 08:16 AM  #6 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty  

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