Solution_to_Problem_of_the_Week_1

# Solution_to_Problem_of_the_Week_1 - Solution to Problem...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online