$A$ boy on a cliff $49\, m$ high drops a stone. One second later,he throws a second stone after the first. They both hit the ground at the same time. With what speed did he throw the second stone?

(12.1 M/S) For the first stone,we are given:
$u = 0\, m\, s^{-1}, h = 49\, m, g = 9.8\, m\, s^{-2}$
Using the equation of motion $S = ut + \frac{1}{2}at^2$,we have:
$49 = 0 \times t + \frac{1}{2} \times 9.8 \times t^2$
$t^2 = \frac{98}{9.8} = 10$
$t = \sqrt{10} \approx 3.16\, s$
So,the first stone takes $3.16\, s$ to reach the ground.
For the second stone,it is thrown $1\, s$ later but hits the ground at the same time. Therefore,the time taken by the second stone is $t_2 = 3.16 - 1 = 2.16\, s$.
Using $S = ut + \frac{1}{2}at^2$ for the second stone:
$49 = u \times 2.16 + \frac{1}{2} \times 9.8 \times (2.16)^2$
$49 = 2.16u + 4.9 \times 4.6656$
$49 = 2.16u + 22.86$
$2.16u = 49 - 22.86 = 26.14$
$u = \frac{26.14}{2.16} \approx 12.10\, m\, s^{-1}$
Thus,the second stone was thrown with a speed of $12.1\, m\, s^{-1}$.