Answer
$3.40\times 10^{-34}m$; $56.5\times 10^{-6}m$
Work Step by Step
We know that the uncertainty in the baseball position is given as
$\Delta y \ge \frac{h}{2\pi(m_{baseball})(\Delta v_y)}$
We plug in the known values to obtain:
$\Delta y\ge \frac{6.63\times 10^{-34}J.s}{2(3.14)(0.15Kg)(0.05)(41m/s)}$
$\implies \Delta y\ge 3.40\times 10^{-34}m$
Now the uncertainty in electron position is given as
$\Delta y\ge \frac{h}{2\pi m_e \Delta v_y}$
We plug in the known values to obtain:
$\Delta y\ge \frac{6.63\times 10^{-34}J.s}{2(3.14)(9.11\times 10^{-31}Kg)(0.05)(41m/s)}$
$\implies \Delta y \ge 56.5\times 10^{-6}m$
