$A$ ball is dropped from a height of $20\, m$. $A$ second ball is thrown downwards from the same height after one second with initial velocity $u$. If both the balls reach the ground at the same time, calculate the initial velocity of the second ball. (Take $g = 10\, m s^{-2}$).

$(15 M S^{-1})$ For the first ball:
$u = 0, h = 20\, m, g = 10\, m s^{-2}, t = ?$
Using the equation of motion $S = ut + \frac{1}{2}at^2$, we have:
$20 = 0 + \frac{1}{2} \times 10 \times t^2$
$20 = 5t^2$
$t^2 = 4$
$t = 2\, s$
For the second ball:
Since the second ball is thrown $1\, s$ later and reaches the ground at the same time, its time of travel is $t' = 2 - 1 = 1\, s$.
$u = ?, h = 20\, m, g = 10\, m s^{-2}, t' = 1\, s$
Using $S = ut' + \frac{1}{2}at'^2$, we have:
$20 = u(1) + \frac{1}{2} \times 10 \times (1)^2$
$20 = u + 5$
$u = 20 - 5 = 15\, m s^{-1}$