anonymous
  • anonymous
I provided a picture of my work to the problem. Part 1) A 47 gram golf ball is driven from the tee at a speed of 48 m/s and rises to a height of 25m. Determine the kinetic energy (in Joules) of the ball at its highest point assuming air resistance is negligible. Part 2) In the previous question, what is the speed of the golf ball (in m/s) when it is 6m below its highest point?
Physics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
@Michele_Laino I know that you had helped me yesterday with this, but I am getting the wrong answer. Do you know where I am going wrong?
matt101
  • matt101
The trouble here is that you can't set kinetic energy equal to potential energy the way you did. At the maximum height, all the kinetic energy contributing to the ball's VERTICAL speed has been converted into potential energy, but there is still kinetic energy contributing to the ball's HORIZONTAL speed (this is what we're trying to solve for). Instead, the equation you need to use is this: \[KE_i=PE_{h_{\max}}+KE_{h_{\max}}\] By conservation of energy, ALL the kinetic energy you start with is equal to the gravitational potential energy at the max height PLUS the kinetic energy at the max height. We're asked to find KE(hmax), so we can rewrite as follows: \[KE_{h_{\max}}=PE_{h_{\max}}-KE_i\]\[KE_{h_{\max}}=mgh_{\max}-\frac{1}{2}mv_i^2\] Plug in the information you have available and solve! For the second question, you can actually use the EXACT same equation because conservation of energy still applies. You will just need to update the values of some variables in the equation based on the new situation. Let me know if you have any questions!

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matt101
  • matt101
I would also add that you need to be careful with units. Always make sure you're working in meters (m), kilograms (kg), and seconds (s). If you're given information with different units, make sure you convert them!
Michele_Laino
  • Michele_Laino
please you have to measure the mass using Kg, not grams, so we have: \(m=0.047\) Kg furthermore, if I simplify the equation which you have used, I get: \[\Large {h_{MAX}} = \frac{{v_0^2}}{{2g}} = \frac{{{{48}^2}}}{{2 \cdot 9.81}} = ...\]

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