તકતીની જાડાઈ $s $ માટે  $v = (19/20)u$

${v^2}\, = \,\,{u^2}\,\, + \,\,2as\,\, \Rightarrow \,\,{\left( {\frac{{19}}{{20}}u} \right)^2}\,\, = {u^2}\, + \,\,2as\,\,\,$

$ \Rightarrow \,\,2as\,\, = \,\,\left( {\frac{{361}}{{400}}} \right){u^2}\, – \,\,{u^2}\,\, \Rightarrow \,\,\,a\,\, = \,\, – \,\,\left( {\frac{{39}}{{400}}} \right)\,\,\,\frac{{{u^2}}}{{2s}}\,$ નો ઉપયોગ કરતાં 

ઓછામાઓછી $ n$  તકતીઓ લેવામાં આવે તો $\,\,{{\text{0}}^{\text{2}}}\,\, = \,\,{u^2}\, + \,\,2a\,(ns)\,$

$\, \Rightarrow \,\,n\,\, = \,\, – \,\,\frac{{{u^2}}}{{2as}}\,\, = \,\,\frac{{{u^2}}}{{2\left( {\frac{{39}}{{400}}} \right)\,\frac{{{u^2}}}{{2s}}\,\, \times \,\,s}}\,\, \approx \,\,10.26$

ઓછામાં ઓછી તકતીઓની સંખ્યા $= 11$