# Air Glider Lab Help

itwasanacident

## Homework Statement

The lab book asks me to graph velocity vs time and use the slope to calculate g, than compare it to the experimental value of g.

The air glider was timed at various distances with a set height. The height was 0.002035m and the distances were 0.20m 0.40m, 0.50m, 0.75m and 1.00m . The average times with respect to distances were ( 0.20m,3.91s) (0.40m,6.11s)(0.50,6.50s)(0.75m,8.22s)(1.00m,9.48s).

## Homework Equations

g = 2S/(t^2 sinx)
instantaneous v(sub i)= (2*y(sub i))/ t(sub i)

## The Attempt at a Solution

I used the first equation to find the experimental value of g, but I have no idea how to work out the velocity. I think im using the wrong y( sub i ). I built a triangle and made the distance (ex: 0.20m) the hypotenuse and used the angle 0.117 degrees; which i got from using 1 meter as the hypotenuse and 0.002035m as the opposite side and using arcsin to find x. The opp ( what i assume is the new height) i get from using hyp*sin x =opp for 0.20m is 0.000408. I assume this is the y (sub i) and i plug this into the equation v (sub i ) = (2*y(sub i))/ t(sub i) using the y i got from the last calculation and the mean time, so v(sub i ) = (2*(0.000408))/3.91 and i get 0.000208. I do this again for each value and i get in order; 0.000208,0.000266,0.000314,0.000372,0.000431 and plot this on excel and the slope i get is 4E-5, this number is no where near 9.8 so im sure im doing something wrong. Im just not sure, any help as soon as possible is appreciated. Thank you

## Answers and Replies

Basic_Physics
What do you mean by set height - always released from the same height? Distances horizontal from the release point or vertical downwards from the point where it was released out of rest or along glide path?

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