Question

a man of height 1.2 m walks away from a lamp hanging at height of 4 m above ground. if man walks with a speed of 2.8m/s determine velocity of the tip of mans shadow.

Answer 1

Answer on Question #53207, Physics / Mechanics | Kinematics | Dynamics

A man of height 1.2 m walks away from a lamp hanging at height of 4 m above ground. If man walks with a speed of 2.8m/s determine velocity of the tip of mans shadow.

Solution:

Let the lamp be at A at height H from the ground, that is AB = H. Let the man be initially at B, below the lamp, his height being equal to BD = h, so that the tip of his shadow is at B. Let the man walk from B to F in time t with speed v, the shadow will go up to C in the same time t with speed v':

$$B F = v t$$ $$B C = v ^ {\prime} t$$

From similar triangles EFC and ABC

$$\frac {F C}{B C} = \frac {E F}{A B} = \frac {h}{H}$$ $$\frac {v ^ {\prime} t - v t}{v ^ {\prime} t} = \frac {h}{H}$$

or

$$v ^ {\prime} = \frac {H v}{H - h} = \frac {4 * 2 . 8}{4 - 1 . 2} = 4 \mathrm {m / s}$$

Answer: $4 \mathrm{~m} / \mathrm{s}$