an electron moving with a velocity of 5*10^6m/s emerges from the sheet of paper of thicness 2.1*10^-4cm with a velocity of 2*10^6m/s.calculate the time taken by the electron to pass through the sheet of paper

Dear student,
Look here we can solve this problem by equations of motion. 
Now, when electron entering the paper sheet its velocity is,v_enter=$5\times {10}^{6}m/s$ and when it emerges from the sheet then the
emerging velocity is v_emerge=$2\times {10}^6 m/s$ and the thickness of the sheet is $s=2·1\times {10}^{-4} cm=2·1\times {10}^{-6} m$
Now to solve this problem we use, v_emerge²= v_enter² + 2fs
$$\begin{gathered} From this equation we can write that,f=\frac{{v_{emerge}}^2-{v_{enter}}^2}{2s}=\frac{4\times {10}^{12}-25\times {10}^{12}}{2\times 2·1\times {10}^{-6}}=-5\times {10}^{18} m/s^{2 } \\ here the negetive sign indicates retardation as the velocity is decrea\sin g after emerging of the paper sheet. \end{gathered}$$
​and now  to solve the problem we use the equation, v_emerge= v_enter+ft
$so, time taken by the electron to cross the paper sheet is, t=\frac{v_{emerge}-v_{enter}}{f}=\frac{-3\times {10}^6}{-5\times {10}^{18}}=6\times {10}^{-13} seconds$
Regards