A man of height 60 cm is walking away from the base of a building at a speed of 2.2 m/s. If the building is 7.2 m of height, then the length of the shadow of man after 5s will be
1 m
Given, AB is the height of building = 7.2 m
EC is the height of man = 0.6 m
Speed of man = 2.2 m/s
∴ Distance travelled in 5 s = Speed × Time = 2.2 m/s × 5s = 11 m
⇒ y = 11 m
In ΔABD and ΔECD,
∠ABD=∠ECD [each 90∘]
∠ADB=∠EDC [common]
∴ΔABD∼ΔECD [by AA similarity]
⇒ABEC=BDCD⇒ABEC−1=BDCD−1⇒AB−ECEC=BD−CDCD⇒7.2−0.60.6=BCCD⇒6.60.6=11CD⇒6.60.6×11=1CD⇒1=1CD⇒CD=1 m
So, the length of the shadow of the man after 5 seconds is 1 m.