Q. 184.5( 2 Votes )

# An object i

For object A

Initial velocity = u = 0m/s

Height = h = 150m

According to 3rd equation of motion

v2 = u2 + 2gh

v2 = 02 + 2(9.8)(150)

v2 = 19.6 × 150

v2 = 2940

v = 54.2m/s

According to 2nd equation of motion

s = ut + 1/2 gt2

s = 0 + 1/2 (9.8)(2)2

s = (9.8)(2)

s = 19.6m

Distance between object and ground = 150 – 19.6 = 130.4m

For object B

Initial velocity = u = 0m/s

Height = h = 100m

According to 3rd equation of motion

v2 = u2 + 2gh

v2 = 02 + 2(9.8)(100)

v2 = 19.6 × 100

v2 = 1960

v = 44.2 m/s

According to 2nd equation of motion

s = ut + 1/2 gt2

s = 0 + 1/2(9.8)(2)2

s = (9.8)(2)

s = 19.6m

Distance between object and ground = 100 – 19.6 = 80.4m

Distance between the two objects after 2s = 130.4 – 80.4

= 50m

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