ન્યુટન ના સૂત્ર પરથી \((\vec v_2-\vec v_1)=e(\vec u_1-\vec u_2)\)

\({\text{v}}\,\, = \,\,\sqrt {{\text{2gh}}} \) અહી, \({{\vec v}_2}\, = \,\,0,\,\,{{\vec u}_2}\, = \,\,0\) (સ્થિર સ્થિતિ આગળ ની સપાટી )

\({{\text{v}}_{\text{1}}}\,\, = \,\,ev\) (વિરુદ્ધ દિશા )\(,\,\,\,\,{\text{v}}\,\, = \,\,\,{{\text{u}}_{\text{1}}}\)

\({v_1}\,\, = \,\,ev\,\,....(\,i)\,,\,\,\,{v_2}\, = \,\,e{v_1}\,\,....(\,\,ii),\,\,\,{v_2}\,\, = \,\,e(ev)\,\,.....(\,\,iii)\,\,\,[(i)\,\& \,(ii)\,\) પરથી \(]\)

\(\therefore \,\,\,\,{v_2}\, = \,\,{e^2}v,\,\,{v_3}\,\, = \,\,{e^3}v,\,\,{v_4}\,\, = \,\,{e^4}v\,.........\,\,{v_n}\, = \,\,{e^n}v\,\,\,\therefore \,\,\,{v_n}\, = \,\,{e^n}\sqrt {2gh} \)