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Kaplunovsky - practice 03

Kaplunovsky - practice 03 - practice 03 Due Feb 5 2007 noon...

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practice 03 – – Due: Feb 5 2007, noon 1 Question 1, chap 2, sect 7. part 1 of 3 5 points The position of a softball tossed vertically upward is described by the equation y = c 1 t c 2 t 2 , where y is in meters, t in seconds, c 1 = 3 . 72 m / s, and c 2 = 5 . 49 m / s 2 . Find the ball’s initial speed v 0 at t 0 = 0 s. Correct answer: 3 . 72 m / s (tolerance ± 1 %). Explanation: Basic Concepts: v = dx dt a = dv dt = d 2 x dt 2 Solution: The velocity is simply the derivative of y with respect to t : v = dy dt = 3 . 72 m / s 2(5 . 49 m / s 2 ) t, which at t = 0 is v 0 = 3 . 72 m / s . Question 2, chap 2, sect 7. part 2 of 3 5 points Find its velocity at t = 1 . 29 s. Correct answer: 10 . 4442 m / s (tolerance ± 1 %). Explanation: Substituting t = 1 . 29 s into the above for- mula for v , we obtain v = 3 . 72 m / s 2(5 . 49 m / s 2 )(1 . 29 s) = 10 . 4442 m / s . Question 3, chap 2, sect 7. part 3 of 3 0 points Find its acceleration at t = 1 . 29 s. Correct answer: 10 . 98 m / s 2 (tolerance ± 1 %). Explanation: The acceleration is the derivative of velocity with respect to time: a = dv dt = 2(5 . 49 m / s 2 ) = 10 . 98 m / s 2 . Question 4, chap 2, sect 6. part 1 of 1 10 points A ball is thrown upward. Its initial ver- tical speed v 0 , acceleration of gravity g , and maximum height h max are shown in the figure below. Neglect: Air resistance. The acceleration of gravity is 9 . 8 m / s 2 . v 0 9 . 8 m / s 2 h max What is its maximum height, h max (in terms of the initial speed v 0 )? 1. h max = v 2 0 2 g correct 2. h max = v 2 0 2 g 3. h max = 5 v 2 0 8 g 4. h max = 3 v 2 0 2 2 g 5. h max = v 2 0 g 6. h max = 3 v 2 0 4 g
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practice 03 – – Due: Feb 5 2007, noon 2 7. h max = 5 v 2 0 2 2 g 8. h max = v 2 0 4 g 9. h max = 3 v 2 0 2 g Explanation: Basic Concept: For constant accelera- tion, we have v 2 = v 2 0 + 2 a ( y y 0 ) . (1) Solution: The velocity at the top is zero. Since we know velocities and acceleration, Eq. 1 containing v , a , y 0 , and y (no time) is the easiest one to use. Choose the positive direction to be up; then a = g and 0 = v 2 0 + 2 ( g ) ( h max 0) or h max = v 2 0 2 g . Question 5, chap 2, sect 6. part 1 of 2 5 points A ball has an initial speed of 10 m / s. The acceleration of gravity is 9 . 8 m / s 2 . t h y 1 . 5 s 10 m / s What will be its position after 1 . 5 s if it is thrown a) down with an initial speed of 10 m / s? Correct answer: 26 . 025 m (tolerance ± 1 %). Explanation: Basic Concepts: This problem involves initial velocities directed both downward and upward, so all downward motion must be neg- ative. Assuming the initial position is h o = 0, the position at any time t is given by h f = v o t 1 2 g t 2 . Let : t = 1 . 5 s , v o = 10 m / s , and g = 9 . 8 m / s 2 . Solution: h 1 = v t 1 2 g t 2 = (10 m / s) (1 . 5 s) 1 2 (9 . 8 m / s 2 ) (1 . 5 s) 2 = 26 . 025 m .
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