Please explain the image problem fully not only answer I need to be explained......... Please........,.,..,.,.

Height of the cliff, h = 240 m

Velocity of Stone 1, y₁= 10 m/s

Velocity of Stone 2, y₂ = 40 m/s

To calculate the time when stone 1 strikes the ground, use Equation of Motion,

            h = ut + ½ gt²

            -240 = 10 × t – ½ × 10 × t²

            5t² -10t -240 = 0

            t² – 2t – 48 = 0

            t = 8 sec and -6 sec

To calculate the time when stone 2 strikes the ground, use Equation of Motion,

            h = ut + ½ gt²

            -240 = 40 × t – ½ × 10 × t²

            5t² -40t -240 = 0

            t = 12 sec

The relative velocity of the two stone = y₂ – y₁ = 40 – 10 = 30 m/s

Therefore, graph of y₂ – y₁ against time is a straight line.

Hence, when the stone strikes the ground, the relative velocity will increase up to 12 seconds as shown in graph (3).