I\'m trying to find the change in time and change in the vertical position in or

Question

I'm trying to find the change in time and change in the vertical position in order to find vertical acceleration. Can you please help me out. On the paper that our professor gave says to find the first change in time value for the first column we need to subtract .0833-0.

Vertical Position vs. Time change in time Part 2: vertical velocity Vy vertical position (y) 109.75 108.25 106.75 101.25 95.25 87.75 78.25 67.25 54.75 39.25 22.75 4.25 change in (y) 0 0.0416 0.0833 0.125 0.1666 0.2083 0.25 0.2916 0.333 0.375 0.416 0.4583

Explanation / Answer

Time, t=0 is taken as reference time,

Now the change in time = time at that instant - reference time,

similarly, position y= 109.75 is taken as reference position,

change in position = position at the instant - reference position.

Now if you want find the vertical velosity with the following formula,

Vertical velocity = cahnge in position / change in position,

similarly, acceleration = vertical velocity / time.

Time change in time position change in position vertical velocity 0 0 109.75 0 0.0416 0.0416 108.25 1.5 0.0833 0.0833 106.75 3 0.125 0.125 101.25 8.5 0.1666 0.1666 95.25 14.5 0.2083 0.2083 87.75 22 0.25 0.25 78.25 31.5 0.2916 0.2916 67.25 42.5 0.333 0.333 54.75 55 0.375 0.375 39.25 70.5 0.416 0.416 22.75 84 0.4583 0.4583 4.25 105.5

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