kinematics (urgent)－thhzpvx的部落格

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kinematics (urgent)

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1 A drowsy cat spots a flowerpot that sails flrst up and then down past an open window. The pot is in view for a total of 0.50 s, and the top-to-bottom height of the window is 2.00 m. How high above the window top does the flowerpot go?ans: 2.34mCan some1 explain what it's asking and show the steps for... 顯示更多 1 A drowsy cat spots a flowerpot that sails flrst up and then down past an open window. The pot is in view for a total of 0.50 s, and the top-to-bottom height of the window is 2.00 m. How high above the window top does the flowerpot go? ans: 2.34m Can some1 explain what it's asking and show the steps for calculating it? 2. A steel ball is dropped from a building's roof and passes a window, taking 0.125 s to fall from the top to the bottom of the window, a distance of 1 .2m.It then falls to a sidewalk and bounces back past the window, moving from bottom to top in 0.125 s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below bottom of the window is 2.00 s. How tall is the building? ans: 20.4m How to do? 圖片參考：http://imgcld.yimg.com/8/n/HA01097219/o/701110010053113873469810.jpg ans(for 69): 0.56m How to do? 3. can some1 briefly tell me what is a differential of a function? thanks 更新: for 1, shouldn't u = 9.225 instead of 6.77? cause if u=9.225, then the ans from the top is 2.34m

最佳解答:

1. The time that the pot passes through the window is 0.5/2 s = 0.25 s Apply equation s = ut + (1/2)at^2 for the upward journey with s = 2m, t = 0.25 s, a =-g(=-9.81 m/s2), u =? hence, 2 = 0.25u + (1/2)(-9.81)(0.25^2) u = 6.77 m/s For the journey from the bottom of the window to the highest point, apply the equation" v^2 = u^2 + 2a.s with v = 0 m/s, u = 6.77 m/s, a = -g, s =? hence, 0 = 6.77^2 + 2.(-9.81)s s = 2.339 m Therefore, the pot reaches (2.339 - 2) m = 0.339 m above the window top. [your given answer is the height from the window bottom] 2. For the journet passing through the window, let U be the velocity at the window bottom, use equation: s = vt-(1/2)gt^2 with s = -1.2 m, v = U, g = -9.81 m/s2, t = 0.125 s hence, -1.2 = 0.125U - (1/2).(-9.81).(0.125^2) U = -10.213 m/s For the jourey from the window bottom to the ground, apply equation:v = u +at with u = -10.213 m/s, a = -9.81 m/s2, t = 1 s, v =? v = [-10.213 + (-9.81) x 1] m/s = -20.023 m/s For the journey from the roof to ground, use equation: v^2 = u^2 + 2a.s with v = -20.023 m/s, u = 0 m/s, a = -9.81 m/s2, s =? (-20.023)^2 = 2.(-9.81).s s = -20.43 m The height of the building is 20.43 m 3. The difference in speed is the difference in areas bounded by the two curves with the time axis. You need to calculate the areas bit by bit. I think "differential" is a mathematical term, not a term in physics. 2011-10-01 23:10:34 補充： sorry...I made a wrong calculation in Q(1). u should be 9.225 m/s, this gives the height reached is 2.34m from top of window.