A body on a 20m high cli drops a stone. One second later, he throws down another stone. Both the stones hit the ground simultaneously. Find the
initial velocity of the second stone g=10m/sec2.

Taking the downward direction as positive

Stone 1:

Initial velocity (u) = 0 m/s

Acceleration (a) = 10 m/s²

Time taken to reach the ground (t) = t seconds

Distance covered (s) = 20 m

Stone 2:

Initial velocity (u) = u m/s

Acceleration (a) = 10 m/s²

Time taken to reach the ground (t) = (t-1) seconds  

[where t is the time taken by stone 1]

Distance covered (s) = 20 m

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Stone 1:

Using the second equation of motion:

s = ut + 1/2*at²

replacing the variables for Stone 1

20 = (0)(t) + 1/2(10)(t)²

20 = 5t²

Dividing both sides by 5

t² = 4

Taking the square root of both the sides

t = 2 seconds

Hence, Stone 1 reaches the ground in 2 seconds

Stone 2:

The time taken by Stone 2 to reach the ground depends on the time taken by stone 1. Since we defined the time taken by stone 2 as:

Time taken by Stone 2 = (Time taken by Stone 1) - 1

replacing the values:

Time taken by Stone 2 = 2 - 1

Time taken by Stone 2 = 1 second

Hence, Stone 2 will reach the ground in 1 second

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According to the second equation of motion:

s = ut + 1/2 at²

replacing the values for Stone 2

20 = (u)(1) + 1/2(10)(1)²

20 = u + 5

u = 15 m/s

Therefore, the Stone 2 is thrown at a velocity of 15 m/s downwards

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