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PHYSICS Projectile motion
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Q1 http://xs.to/xs.php?h=xs232&d=08416&f=img035313.jpg 淨係做part d 就可以=] Q2 http://xs.to/xs.php?h=xs232&d=08416&f=img036147.jpg 識既朋友唔該教教我>3< thank you~~~ 更新: 首先多謝Audrey Hepburn你咁快答左我 但我小小唔明啊... 第1題既 Final K.E. = 1/2 m(3.75 + 3.52) = 8m J 唔係好明呢個= =? where does 3.52 come from?
1.d. The kinetic energy of the coin when it enters the dish is smaller than that just leave the contestant's hand. It is because some of the initial K.E. of the coin is converted to the G.P.E. gained by the coin. Let m be the mass of the coin Initial K.E. = 1/2 mv2 = 1/2 m(7)2 = 24.5m J For the coin just enter the dish, for the vertical component By v2 = u2 + 2as v2 = (7sin60*)2 + 2(-10)(1.65) v2 = 3.75 Final K.E. = 1/2 m(3.75 + 3.52) = 8m J which is smaller than the initial K.E. 2.a. Consider the vertical motion, By v2 = u2 + 2as 0 = (4.88sin35*)2 + 2(-10)s s = 0.39 m So, the maximum height = 1.83 + 0.39 = 2.22 m b. Consider the vertical motion, take upward be positive direction. s = ut + 1/2 at2 -1.83 = (4.88sin35*)t + 1/2 (-10)t2 Time of flight, t = 0.946 s Consider the horizontal motion, the ball is moving with uniform horizontal speed So, the required distance = (4.88cos35*)(0.946) = 3.78 m c. Let the initial speed be u Horizontally, utcos35* = 4.21 t = 4.21 / ucos35* ─── (1) Vertically, 3.05 - 1.83 = utsin35* + 1/2 (-10)t2 1.22 = usin35*(4.21 / ucos35*) - 5(4.21 / ucos35*)2 (1.22 - 4.21tan35*)u2 + 88.6205/cos235* = 0 u = 8.73 ms-1 d. Vertically, v2 = u2 + 2as 0 = (8.73sin35*)2 + 2(-10)s s = 1.25 m Max. height = 1.25 + 1.83 = 3.08 m By v = u + at 0 = 8.73sin35* - 10t Time of flight, t = 0.50 s So, the horizontal distance from the basket = 4.21 - 8.73(0.50)cos35* = 0.63 m
其他解答:
PHYSICS Projectile motion
發問:
Q1 http://xs.to/xs.php?h=xs232&d=08416&f=img035313.jpg 淨係做part d 就可以=] Q2 http://xs.to/xs.php?h=xs232&d=08416&f=img036147.jpg 識既朋友唔該教教我>3< thank you~~~ 更新: 首先多謝Audrey Hepburn你咁快答左我 但我小小唔明啊... 第1題既 Final K.E. = 1/2 m(3.75 + 3.52) = 8m J 唔係好明呢個= =? where does 3.52 come from?
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最佳解答:1.d. The kinetic energy of the coin when it enters the dish is smaller than that just leave the contestant's hand. It is because some of the initial K.E. of the coin is converted to the G.P.E. gained by the coin. Let m be the mass of the coin Initial K.E. = 1/2 mv2 = 1/2 m(7)2 = 24.5m J For the coin just enter the dish, for the vertical component By v2 = u2 + 2as v2 = (7sin60*)2 + 2(-10)(1.65) v2 = 3.75 Final K.E. = 1/2 m(3.75 + 3.52) = 8m J which is smaller than the initial K.E. 2.a. Consider the vertical motion, By v2 = u2 + 2as 0 = (4.88sin35*)2 + 2(-10)s s = 0.39 m So, the maximum height = 1.83 + 0.39 = 2.22 m b. Consider the vertical motion, take upward be positive direction. s = ut + 1/2 at2 -1.83 = (4.88sin35*)t + 1/2 (-10)t2 Time of flight, t = 0.946 s Consider the horizontal motion, the ball is moving with uniform horizontal speed So, the required distance = (4.88cos35*)(0.946) = 3.78 m c. Let the initial speed be u Horizontally, utcos35* = 4.21 t = 4.21 / ucos35* ─── (1) Vertically, 3.05 - 1.83 = utsin35* + 1/2 (-10)t2 1.22 = usin35*(4.21 / ucos35*) - 5(4.21 / ucos35*)2 (1.22 - 4.21tan35*)u2 + 88.6205/cos235* = 0 u = 8.73 ms-1 d. Vertically, v2 = u2 + 2as 0 = (8.73sin35*)2 + 2(-10)s s = 1.25 m Max. height = 1.25 + 1.83 = 3.08 m By v = u + at 0 = 8.73sin35* - 10t Time of flight, t = 0.50 s So, the horizontal distance from the basket = 4.21 - 8.73(0.50)cos35* = 0.63 m
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