Question

A ball is dropped vertically downwards and after 1 second another ball is dropped from the same height what is distance between the two balls 3 seconds after the first ball is

Answer 1

distance travelled by the first ball in 3 sec after it is dropped is

s = u t + 1/2 g t^2

or, s 1 = 0 + 1/2 × 10 × 3^2 = 45 m

distance travelled by the second ball in 2 sec after it is dropped after 1 sec of the first ball

s 2 = u t + 1/2 g t^2 = 0 + 1/2 × 10 × 2^2

= 20 m

distance between the two balls 3 seconds after the first ball is dropped is

s 1 - s 2 = 45 - 20 = 25 m

Answer 2

☆$\red {HELLO\:DEAR}$☆

Distance travelled by the first ball in 3 sec after it is dropped is:-

s = u t + 1/2 g t^2

or, s 1 = 0 + 1/2 × 10 × 3^2 = 45 m

Distance travelled by the second ball in 2 sec after it is dropped after 1 sec of the first ball:-

s 2 = u t + 1/2 g t^2 = 0 + 1/2 × 10 × 2^2

= 20 m

Distance between the two balls 3 seconds after the first ball is dropped is:-

s 1 - s 2 = 45 - 20 = 25 m

☆$\green {VISHU\:PANDAT}$☆

☆$\blue {FOLLOW\:ME}$☆