Experiment 6 Questions

Question:

I have been working on experiment six, and I am to the part with the equation 'ds =.5g 'dt^2. Could you explain to me what is suppose to fit into the equation. And what I am trying to find.

Answer:

(Observation:  Your work has been very good, but from this and some of your previous questions (all excellent questions), it seems that you might be getting a little bit ahead of the problem sets with your experiments. Be sure you thoroughly understand the Introductory Problems through Set 2, all of which were assigned prior to Experiment 6.)

In the equation, `ds stands for displacement (distance with a direction--i.e., `ds can be positive or negative so we call it displacement instead of distance), `dt for the time interval and g for the acceleration of gravity, which is 9.8 m/s/s.

• So if the ball falls .15 meters from an initial vertical velocity of 0 (it could have any horizontal velocity; vertical velocity is 0 if it is neither rising nor falling), we would have `ds = .15 m and g = 9.8 m/s/s.

• Solving `ds = .5 g `dt^2 for `dt we would obtain `dt = `sqrt( 2 `ds / g) = `sqrt(2 * .15 meters / (9.8 m/s^2) ) = `sqrt(.03 s^2) = .54 sec (approx.).

• Note that the `sqrt indicates 'square root of'.