Answer
$11~m/s$
Work Step by Step
Let the positive direction be upward and the ground be the origin.
First, we will find the velocity of the the first chestnut when it has fallen $2.5~m$.
$v_0=0$, $a=-g=-9.81~m/s^2$, $x_0=10~m$, $x=7.5~m$
$\Delta x=x-x_0=-2.5~m$
$v^2=v_0^2+2a\Delta x$
$v^2=0^2+2(-9.81~m/s^2)(-2.5~m)$
$v=±\sqrt {2(-9.81~m/s^2)(-2.5~m)}=±7.00~m/s$
$v=-7.00~m/s$, because the chestnut is moving downward.
Now, let the initial time, $t=0$, be the moment the second chestnut is thrown downward. Let's find the time it takes for the first chestnut to reach the ground.
$x_{1_0}=7.5~m$, $x_1=0$, $a=-g=-9.81~m/s^2$
$x_1=x_{1_0}+v_{1_0}t+\frac{1}{2}at^2$
$0=7.5~m+(-7.00~m/s)t+\frac{1}{2}(-9.81~m/s^2)t^2$
$(4.905~m/s^2)t^2+(7.00~m/s)t-7.5~m=0$
This is a quadratic equation for t.
$t=\frac{-(7.00~m/s)±\sqrt {(7.00~m/s)^2-4(4.905~m/s^2)(-7.5~m)}}{2(4.905~m/s^2)}$
$t_1=0.714~s$
$t_2=-2.141~s$ (this solution is not valid)
That is, it takes $0.714~s$ to the first chestnut to reach the ground. But, both nuts reach the ground at the same time.
$x_{2_0}=10~m$, $x_2=0$, $a=-g=-9.81~m/s^2$, $t=0.714~s$
$x_2=x_{2_0}+v_{2_0}t+\frac{1}{2}at^2$
$0=10~m+v_{2_0}(0.714~s)+\frac{1}{2}(-9.81~m/s^2)(0.714~s)^2$
$v_{2_0}=\frac{-10~m+\frac{1}{2}(9.81~m/s^2)(0.714~s)^2}{0.714~s}=-11~m/s$
$speed=|v|=11~m/s$
