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I just did a basic projectile motion experiment. What I did was build a small catapult out of small Popsicle sticks and rubber bands, and then I used it to launch a small ball at an angle above the ground. I wanted to figure out the total potential energy of the catapult when I pulled it back to launch the ball.

To do this, I figured that I should find the ball's total energy in both the x- and y- axis (ignoring anything in the z-axis). For the x-axis, I figured out the ball's horizontal Ek by dividing the horizontal distance it traveled by the time it took. This gave me the ball's horizontal speed, which I used to find its Ek. Due to the conservation of energy, I reasoned that the ball's energy in the x-axis must have been this value for the entire time (there were no horizontal forces acting on the ball, ignoring drag).

For the y-axis, I measured the distance above the ground that its highest point of flight was. Then I figured out the GPE at the highest point by using this value. Once again, I reasoned that the ball's energy at all times in the y-axis was this value, due to the conservation of energy (once again, ignoring drag).

At this point, I had the ball's energy in the x- axis and y-axis. To find the ball's total energy, I just added up both of the values. This told me how much potential energy was in the catapult when it was drawn back.

I was wondering if my steps were correct. Was what I did with adding up the energy in the x- and y- axis to find the total energy correct? I figured this was allowed since energy is a scalar, so direction shouldn't matter.

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  • $\begingroup$ Energy is not a vector. It does not have components. (Occasionally you find cases where you can treat it as if it did, but those are the exception, and you have to understand how energy works as a scalar to be able to identify them. However, this is not such a case.) $\endgroup$ – dmckee Apr 15 '17 at 1:05
  • $\begingroup$ Thanks. I wasn't really thinking of energy as having components, it was just a way for me to work with it easier. Was my method or finding total energy correct? $\endgroup$ – Inertial Ignorance Apr 15 '17 at 1:37
  • $\begingroup$ You can find the launch speed from the maximum range, without any need to measure time of flight. $\endgroup$ – sammy gerbil Apr 15 '17 at 4:33
  • $\begingroup$ Isn't it necessary to measure the time the projectile took to reach that distance though? Since speed = distance / time. $\endgroup$ – Inertial Ignorance Apr 15 '17 at 5:10
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Talking about kinetic energy along the x- or y-axis is not quite correct, but close enough if you calculate it explicitly like this: $$K = \frac{1}{2}mv^2 = \frac{1}{2}m(v_x^2 + v_y^2) = \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2$$

Otherwise, your calculations are correct. An easier method requiring only one measurement is to fire the projectile straight up while timing how long it's in the air. From this you can get the maximum height and thus the total energy.

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  • $\begingroup$ Thanks for your answer - yes, firing it straight up would definitely be more straightforward. I mainly just thought of energy as x- and y- components in order to make my calculations clearer. Did my calculations arrive at the exact answer, or do you think I was a bit off? $\endgroup$ – Inertial Ignorance Apr 15 '17 at 4:00
  • $\begingroup$ The calculation was correct. By "not quite correct," I only meant that talking about "horizontal kinetic energy" is an unusual way of talking about kinetic energy. This is because kinetic energy does not have a direction like other vectors like velocity, so dividing it into horizontal and vertical components will cause some readers to raise their eyebrows. You could phrase it more clearly as "the total kinetic energy is the sum of the kinetic energy due to horizontal motion and vertical motion." $\endgroup$ – Mark H Apr 15 '17 at 4:07

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