Answer to Question #244506 in Quantum Mechanics for Ashafa

Question

1. The force required to draw back the string of an archers bow by a distance 'x' from it equilibrium position is given by f=kx2 if all the energy stored in the bow is communicated to the arrow, show that the maximum range of the arrow is given by R=2kx2/3mg

Answer 1

$F=kx^2,$

$A=\int Fdx=\frac{kx^3}3,$

$E=\frac{mv^2}2,$

$A=E,\implies v^2=\frac{2kx^3}{3m},$

$l_{max}=\frac{v^2}g=\frac{2kx^3}{3mg}=R.$