Asked by Teri
a light shines from the top of a pole 50ft high. a ball is dropped from the same height from a point 30 ft away from the light. how fast is the ball's shadow moving along the ground 1/2 sec later? (assume the ball falls at a distance of s=16t^2 in t sec)
Answers
Answered by
Steve
when the ball is at height h, the shadow is at distance x from the base of the light pole. Using similar triangles,
x/50 = (x-30)/h
h = 50(x-30)/x = 50 - 1500/x
now, h = 50-s = 50-16t^2, so
dh/dt = -32t, and
h(1/2) = 50 - 16(1/4) = 46
x/50 = (x-30)/46, so x=375
-32t = 1500/x^2 dx/dt
at t = 1/2,
-16 = 1500/(140625) dx/dt
dx/dt = -1500
so, the shadow is moving toward the pole at 1500 ft/s
x/50 = (x-30)/h
h = 50(x-30)/x = 50 - 1500/x
now, h = 50-s = 50-16t^2, so
dh/dt = -32t, and
h(1/2) = 50 - 16(1/4) = 46
x/50 = (x-30)/46, so x=375
-32t = 1500/x^2 dx/dt
at t = 1/2,
-16 = 1500/(140625) dx/dt
dx/dt = -1500
so, the shadow is moving toward the pole at 1500 ft/s
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