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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 windows top A and v B its speed as it passes the windows bottom B . How much did the ball speed up as it passed the window; i.e. , calculate 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 = v t = v B v A t and the change of its velocity during time t AB is simply vectorv = vectorv B vectorv A = g t AB , assuming the downward direction to be positive. 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 windows 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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 Fall '09
 MIHALISIN
 Physics

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