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1. A stone is thrown upward from a cliff 100m above the sea level. The maximum height reached by the stone is 18m above the cliff. < You must take downwards as positive. >a) Find the initial velocity of the stone.b) Find the time taken for the stone to travel from the cliff to the sea level.c)... 顯示更多 1. A stone is thrown upward from a cliff 100m above the sea level. The maximum height reached by the stone is 18m above the cliff. < You must take downwards as positive. > a) Find the initial velocity of the stone. b) Find the time taken for the stone to travel from the cliff to the sea level. c) Find the velocity at which the stone hit the sea. d) Sketch an acceleration-time graph. e) Sketch a velocity-time graph of the stone. f) Sketch a displacement-time graph of the stone.
最佳解答:
a) s=-18m, v=o , a= 10 By v^2 = u^2 =2as -u^2 =2(10)(-18) u =19 so, the inital velocity of the stone is 19ms^-1.(upward) b) u=-19, a=10 , s=-100 by s=ut+0.5at^2 -100= -19t+5t^2 5t^2-19t-100=0 t=6.759 or t=-2.959(rejected) so, the time taken for the stone to travel from the cliff to the sea level is 6.76s. c) u=-19 a=10 s=100 by v^2 =u^2 +2as v^2 = -19^2 + 2(10)(100) v =48.6 so, the velocity at which the stone hit the sea is 48.6ms^-1.
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發問:1. A stone is thrown upward from a cliff 100m above the sea level. The maximum height reached by the stone is 18m above the cliff. < You must take downwards as positive. >a) Find the initial velocity of the stone.b) Find the time taken for the stone to travel from the cliff to the sea level.c)... 顯示更多 1. A stone is thrown upward from a cliff 100m above the sea level. The maximum height reached by the stone is 18m above the cliff. < You must take downwards as positive. > a) Find the initial velocity of the stone. b) Find the time taken for the stone to travel from the cliff to the sea level. c) Find the velocity at which the stone hit the sea. d) Sketch an acceleration-time graph. e) Sketch a velocity-time graph of the stone. f) Sketch a displacement-time graph of the stone.
最佳解答:
a) s=-18m, v=o , a= 10 By v^2 = u^2 =2as -u^2 =2(10)(-18) u =19 so, the inital velocity of the stone is 19ms^-1.(upward) b) u=-19, a=10 , s=-100 by s=ut+0.5at^2 -100= -19t+5t^2 5t^2-19t-100=0 t=6.759 or t=-2.959(rejected) so, the time taken for the stone to travel from the cliff to the sea level is 6.76s. c) u=-19 a=10 s=100 by v^2 =u^2 +2as v^2 = -19^2 + 2(10)(100) v =48.6 so, the velocity at which the stone hit the sea is 48.6ms^-1.
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