標題:

physics projectile motion

發問:

1. An object is projected so that it just clears two obstacles each 25m high, which are situated 160m from each other. if the time of passing between the obstacles is 2.5s, calculate(a) the full range of projection(b) the initial velocity of projection of the object.2. A body is projected from the... 顯示更多 1. An object is projected so that it just clears two obstacles each 25m high, which are situated 160m from each other. if the time of passing between the obstacles is 2.5s, calculate (a) the full range of projection (b) the initial velocity of projection of the object. 2. A body is projected from the top of a tower 25m high with a velocity of 40m s ^-1 at an angle of elevation 30 degree with the horizontal. find (a) the time for it to strike the ground (b) the distance from the foot of the tower when the impact occurs. (c) the angle at which the body strikes the ground.

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1. Consider the horizontal motion, the projectile is moving with constant horizontal speed. s = u't 160 = u'(2.5) Horizontal speed, u' = 64 ms-1 Consider the vertical motion, the vertical displacement of the projectile through the two obstacles is zero. Take upward direction be positive. So, s = vt + 1/2 gt2 0 = v(2.5) + 1/2 (-10)(2.5)2 Vertical velocity when passing through the first obstacle, v = 12.5 ms-1 Now, By v2 = u2 + 2gs (12.5)2 = u2 + 2(-10)(25) Initial vertical velocity, u = 25.62 ms-1 So, now, we can find out the time of flight for the whole journey. Let it be T by s = uT + 1/2 gT2 0 = (25.62)T + 1/2 (-10)T2 Time of flight, T = 5.12 s So, the full range, R = u'T = 64(5.12) = 328 m b. Initial velocity = (u'2 + u2)1/2 = [(64)2 + (25.62)2]1/2 = 68.9 ms-1 Let x be the angle of the initial velocity with the horizontal tanx = 25.62 / 64 x = 21.8* So, the initial velocity is 68.9 ms-1 with a direction 21.8* with the horizontal. 2.a. Consider the vertical motion, let upward direction be positive. By s = ut + 1/2 gt2 -25 = (40sin30*)t + 1/2 (-10)t2 t2 - 4t - 5 = 0 Time, t = 5 s or -1 (rejected) So, the time required is 5 s. b. Consider the horizontal motion, the projectile is moving with constant horizontal speed. So, R = ut R = 40cos30* (5) Distance required, R = 173 m c. Consider the vertical motion, By v2 = u2 + 2gs v2 = (40sin30*)2 + 2(-10)(-25) Vertical velocity, v = -30 ms-1 Let y be the angle of the body strike the ground with respect to the horizontal. tany = 30 / 40cos30* y = 40.9* So, the body strikes the ground with an angle of 40.9* with the horizontal. 2008-09-28 19:55:57 補充： 第一題首先考慮projectile由第一個obstacle去第二個的旅程 由於兩者高度一樣，均為25 m，所以在這個旅程中，vertical displacement為0

其他解答:

thanks for yr answer 我想問點解第1題既vertical displacement係zero?