<h3><u>
We are given:</u></h3>
<em>Taking the downward direction as positive</em>
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
<em>[where t is the time taken by stone 1]</em>
Distance covered (s) = 20 m
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<h3><u>
Time taken by the Stones to reach the ground:</u></h3>
Stone 1:
Using the second equation of motion:
s = ut + 1/2*at²
<em>replacing the variables for Stone 1</em>
20 = (0)(t) + 1/2(10)(t)²
20 = 5t²
<em>Dividing both sides by 5</em>
t² = 4
<em>Taking the square root of both the sides</em>
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
<em>replacing the values:</em>
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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<h3><u>
Initial Velocity of Stone 2:</u></h3>
According to the second equation of motion:
s = ut + 1/2 at²
<em>replacing the values for Stone 2</em>
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