Unformatted text preview: Solution to Problem of the Week #1 a. The average velocity as the ball travels between location A and B: ( x B − x A ) ( yB − yA ) ( z B − z A ) vaverage =
,
,
= 22.3, 26.1, 0 m / s Δt AB
Δt AB
Δt AB b. If the ball continued to travel at the same average velocity during the next second, where would it be at t = 2 seconds? xC = x B + vaverage, x ⋅ Δt BC = 22.3 + 22.3 ⋅ 1 = 44.6 m
yC = yB + vaverage, y ⋅ Δt BC = 26.1 + 26.1 ⋅ 1 = 52.2 m The predicted coordinates at t = 2 seconds is therefore: 44.6, 52.2, 0 m c. How does your prediction from part (b) compare to the actual position of the ball at location C? Let’s calculate the difference between the actual position and the predicted position. xC , yC , zC actual − xC , yC , zC predicted = ( 40.1 − 44.6 ), ( 38.1 − 52.2 ), 0 = −4.5, −14.1, 0 m In the absence of air resistance, we would expect that our predicted xC and the actual xC would agree since in this case gravity is the only force acting and it acts in the y
direction. Instead, we see that air resistance affects our prediction of xC. We also expect that our prediction of the y
coordinate is affected by both the force due to gravity and the drag force due to the atmosphere. ...
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 Fall '08
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 0 m, the00, vaverage, 52.2 m, 44.6 m

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