A man of height h walks in a straight path towards a lamp post of height H with velocity v. Then velocity of the edge of the shadow on the ground will be
A
hv(H−h)
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B
Hv(H−h)
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C
hv(H+h)
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D
(H−h)hv
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Solution
The correct option is BHv(H−h) Let the situation be as shown below
Here, OL is the lamp post, MP is the man who is walking towards post with the velocity v. x is the length of the shadow and y is the position or distance of man from the post.
So from geometry, ΔLOSandΔMPS are similar ⇒Hx=h(x−y)⇒(H−h)x=Hy
Differentiating L.H.S and R.H.S w.r.t time, we get ⇒(H−h)dxdt=Hdydt
So, let velocity of edge of the shadow be vs which is dxdt
and, dydt is the velocity of man i.e. v ∴ we have (H−h)vs=Hv⇒vs=Hv(H−h)