Question

Find the ratio of the distances travelled by a freely falling body in first,second and third second of its fall.

Answer 1

we know that

s = ut + (1/2)at²

in the first case

now, here u = 0

t₁ = 1 s

a = 9.8 m/s²

so,

s₁ = (1/2)x9.8x1²

or

the distance travelled after 1st second will be

s₁ = 4.9 m

here, the velocity after 1st second will be

v₁ = at₁

or

v₁ = 9.8x1 = 9.8 m/s

now, in the second case

s₂ = v₁t₂+ (1/2)at₂²

here

v₁ = 9.8 m/s

a = 9.8 m/s²

t₂ = 1 s

thus,

s₂ = 9.8x1 + (1/2)x9.8x1²

or

the distance travelled after 2nd second will be

s₂ = 14.7 m

the velocity after the 2nd second will be

v₂ = v₁ + at₂

or

v₂ = 9.8 + 9.8x1

thus,

v₂ = 19.6 m/s

similarly,

the distance travelled after 3rd second will be

s₃ = v₂t₃ + (1/2)at₃²

here

v₁ = 19.6 m/s

a = 9.8 m/s²

t₃ = 1 s

or

the distance travelled after 3rd second will be

s₃ = 19.6 m

thus, the ratio will be

s₁:s₂:s₃ = 4.9:14.7:19.6 = 1:3:4