Two balls are thrown horizontally, one from the window of first floor which is $3 \mathrm{~m}$ high from the ground and second from the second floor which is $6 \mathrm{~m}$ high from the ground, of a multi storey building, with the same speed of $6 \mathrm{~ms}^{-1}$. Calculate the distance that will separate the two balls when they hit the ground.

The time taken for a ball thrown horizontally from a height $h$ to reach the ground is given by $t = \sqrt{\dfrac{2h}{g}}$.

For the first ball thrown from height $h_1 = 3$ m:
$t_1 = \sqrt{\dfrac{2 \times 3}{9.8}} = \sqrt{\dfrac{6}{9.8}} \approx 0.7825$ s.

The horizontal distance covered by the first ball is $x_1 = v \times t_1 = 6 \times 0.7825 = 4.695$ m.

For the second ball thrown from height $h_2 = 6$ m:
$t_2 = \sqrt{\dfrac{2 \times 6}{9.8}} = \sqrt{\dfrac{12}{9.8}} \approx 1.1066$ s.

The horizontal distance covered by the second ball is $x_2 = v \times t_2 = 6 \times 1.1066 = 6.6396$ m.

The distance separating the two balls when they hit the ground is the difference in their horizontal ranges:
$\Delta x = x_2 - x_1 = 6.6396 - 4.695 = 1.9446$ m.

Rounding to two decimal places, the distance is $1.94$ m.

Answer: 1.94 m