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Unformatted text preview: patel (kp9583) – HW2 – Mackie – (20211) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. DEADLINE IS CENTRAL TIME! 001 10.0 points An object is released from rest on a planet that has no atmosphere. The object falls freely for 3 . 78 m in the first second. What is the magnitude of the acceleration due to gravity on the planet? Correct answer: 7 . 56 m / s 2 . Explanation: Let : s = 3 . 78 m . s = 1 2 a t 2 a = 2 s t 2 = 2 (3 . 78 m) (1 s) 2 = 7 . 56 m / s 2 . 002 (part 1 of 2) 10.0 points A ball is dropped from rest at point O (height unknown). After falling for some time, it passes by a window of height 2 . 9 m and it does so during time t AB = 0 . 4 s. The acceleration of gravity is 9 . 8 m / s 2 . O A B 2 . 9 m b b b b b b b b b b b x y The ball accelerates all the way down; let v A be its speed as it passes the window’s top A and v B its speed as it passes the window’s bottom B . How much did the ball speed up as it passed the window; i.e. , calculate Δ v down = v B − v A ? Correct answer: 3 . 92 m / s. Explanation: Let : h = 2 . 9 m , t AB = 0 . 4 s , and g = 9 . 8 m / s 2 . O A B h t AB b b b b b b b b b b b b t y Assume: Down is positive. The ball falls under a constant acceleration g , so g = Δ v t = v B − v A t and the change of its velocity during time t AB is simply Δ vectorv = vectorv B − vectorv A = g t AB , assuming the downward direction to be positive. Δ v down = (9 . 8 m / s 2 )(0 . 4 s) = 3 . 92 m / s . 003 (part 2 of 2) 10.0 points Calculate the speed v A at which the ball passes the window’s top. Correct answer: 5 . 29 m / s. Explanation: Down is positive. Given the uniform down ward acceleration g , we have y ( t ) = y ( t A ) + vectorv A ( t − t A ) + 1 2 g ( t − t A ) 2 and hence h = y ( t B ) − y ( t A ) = vectorv A ( t B − t A ) + 1 2 g ( t B − t A ) 2 = vectorv A t AB + 1 2 g t 2 AB . patel (kp9583) – HW2 – Mackie – (20211) 2 Solving this equation for the velocity vectorv A , we obtain vectorv A = h t AB − g t AB 2 = 2 . 9 m . 4 s − (9 . 8 m / s 2 )(0 . 4 s) 2 = 5 . 29 m / s . Thus the speed v A = bardbl vectorv A bardbl = 5 . 29 m / s. 004 10.0 points You are trying to find out how high you have to pitch a water balloon in order for it to burst when it hits the ground. You discover that the balloon bursts when you have pitched it to a height of 16 m....
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This note was uploaded on 12/10/2009 for the course PHYSICS 2021 taught by Professor Mihalisin during the Fall '09 term at Temple.
 Fall '09
 MIHALISIN
 Physics

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