v2ae 2kx gk upward motion v2gk be2kx downward motion use above result show when bullet hits

V^2=Ae^-2kx – g/k (upward motion) V^2=g/k – Be^2kx (downward motion) Use the above result to show that , when the bullet hits the ground on its return, the speed will be equal to the expression Vo Vt / (Vo^2 + Vt^2)^0.5 In which Vo is the initial upward speed and Vt= (mg/C2)^0.5= terminal speed = (g/k)^0.5 This is result allows one to find the fraction of the initial kinetic energy lost through air friction

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