Question 1094286: A light is at the top of a pole 80m high. A ball is dropped from the same height (80m) from a point 20m from the light.
Assuming that the ball falls according to the law s=16t^2, how fast is the shodow of the ball moving along the ground one second later?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A light is at the top of a pole 80m high. A ball is dropped from the same height (80m) from a point 20m from the light.
Assuming that the ball falls according to the law s=16t^2, how fast is the shodow of the ball moving along the ground one second later?
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s = at^2/2 = 16t^2
ds/dt = 32t --> the ball is falling at 32 ft/sec
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The distance from the light pole to the shadow of the ball is 50 ft
s/20 = 80/x
(using similar triangles).
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Find the distance to the shadow in terms of the position of the ball, call it x.
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s/20 = 80/x
x = 1600/s
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dx/dt = -1600/(s^2)ds/dt
dx/dt = (-1600/(16^2))*16
dx/dt = -100 ft/sec (negative since it's decreasing)
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Oops. It's 80 meters, not 80 feet.
You do the conversions.
Or use 5 m/sec/sec for the acceleration due to gravity.
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