Here. AB = x : distance covered by the edge of the shadow in time t BC = y : distance covered by the man in lime t In ΔACE, tan θ =
CE
AB+BC
=
H
z+y
Similarly in ΔEFD, tan θ =
ED
FD
=
H−h
Y
Since the angle ’θ' is same, we have
H
x+y
=
H−h
y
⇒ Hy = Hx + Hy - xh - yh ⇒ yh = Hx - xh ⇒ yh = x (H - h) ⇒
x
y
=
h
H−h
... (i) Let v be the velocity of the edge of the shadow on the ground. Then x = vt ... (ii) and y = ut ... (iii) substituting the values from eqs (ii) and (iii) in eq (i) we get