Question

a boy of height 150cm is walking away from the base of a lamp post at a speed of 1.6m/s.if the lamp is 4.5cm above the ground find the length of his shadow after 5seconds

Answer 1

in the question the height of the lamp would be probably 4.5 m=450 cm
distance =speed*time
distance travelled
by the boyafter
5 seconds=1.6*5=8 metres=800 cm
length of the shadow=800-450=350cm=3.5 metres
please mark my answer as brainliest
hope it helps............

Answer 2

AB is the height of Lamp post

DE is the height of the boy.

EC is the length of shadow.

BE is the distance covered by the boy in 5 sec= 1.6 m/s × 5s = 8 m

In Δ ABC & Δ DEC:-

1) ∠ ABC  & ∠DEC = 90° ( lamp-post and the boy are perpendicular to the ground ).

2) ∠ ACB is common for both Δ.

∴ ΔABC & ΔDEC are similar Δ.

AB/ DE= BC/ EC

$\frac{4.5}{1.5}$  = BE + EC/EC

3= 8 + EC/ EC

3EC = 8 + EC

3EC -EC = 8

2EC=8

EC= 4m

Hence the length of his shadow after 5seconds will be 4m.

Ans :-  The length of his shadow after 5seconds will be 4m.

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