HardPhysicsfluid
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1. A person of height 1.8 m is walking away from a lamp post of height 5 m along a straight path on the flat ground. If the speed of the person is 70 cm s⁻¹, the speed of the tip of the person's shadow on the ground with respect to the person is
A
0.52 m/s
B
0.68 m/s
C
0.76 m/s
D
0.40 m/s
Explanation
Let H=5m, h=1.8m. By similar triangles, x/h = (x+y)/H where y is person's distance and x is shadow length. x(H-h) = hy. Differentiating: v_shadow = [h/(H-h)] * v_person. v_person = 0.7 m/s. v_shadow = [1.8/3.2] * 0.7 = 0.393 ≈ 0.40 m/s.
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