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A ball is dropped from a boat so that it strikes the surface of a lake with a speed of 16.5 fps. While in the water the ball experiences an acceleration of a=10−0.8v, where a and v are expressed in fps^2 and fps respectively. Knowing the ball takes 3 sec to reach the bottom of the lake, determine the following:

a. the depth of the lake
b. the speed of the ball when it hits the bottom of the lake

1 Answer

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a = 10 - 0.8vv = d/ta = change in velocity / change in time
v1 = 16.5 fps, t1 = 0, d1 = 0 v2 = y fps , t2 = 3, d2 = X
10 - 0.8(d2/3) = (16.5 - 0)/ (0-3)10 - 0.8(d2/3) = 16.5/-3d2 = 58.125 feet
v2 = 58.125/3 v2 = 19.375 fps
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