Motorcycle Safety

First Use Conservation of Kinetic Energy to Determine the Persons Velocity as he is Projected off the Motorcycle: $KE_{ i }=KE_{ f }\\ \quad \frac { 1 }{ 2 } (M+m)v^{2}=\frac { 1 }{ 2 } mv'^{ 2 }\\ \quad \quad v\sqrt{\frac { (M+m) }{ m } }=v'$ Then determine the amount of time the person is in the air: $\\ y_{ f }=y_{ i }+v_{ i }t+\frac { at^{ 2 } }{ 2 } \\ \quad a=-g;v_{ i }=v'sinθ;y_{ i }y_{ f }=0\\ \quad \quad 0=0+v'tsin⁡θ-\frac { gt^{ 2 } }{ 2 } \\ \quad \quad \quad v'sin⁡θ=\frac { gt }{ 2 } \\ \quad \quad \quad \quad t\quad =\quad \frac { 2v'sin⁡θ }{ g }$ Now determine the horizontal distance of the person from the motorcycle: $\\ x_{ f }=x_{ i }+v_{ i }t+\frac { at^{ 2 } }{ 2 } \\ \quad v_{ i }=v'cos⁡θ;x_{ i }a=0\\ \quad \quad x_{ f }=0+v'cos⁡θ(\frac { 2v'sin⁡θ }{ g } )+\frac { 0t^{ 2 } }{ 2 } \\ \quad \quad \quad x_{ f }=\frac { v'^{ 2 }sin⁡2θ }{ g }$ Finally, Plug in all known variables and solve then simplify: $\\ θ\quad =\quad \frac { π }{ 4 } \quad =\quad 45°\\ \quad x_{ f }=\frac { v'^{ 2 }sin\cfrac { π }{ 2 } }{ g } \\ \quad \quad x_{ f }=\frac { (v\sqrt{\frac { (M+m) }{ m } } )^{ 2 } }{ g } \\ \quad \quad \quad x_{ f }=\frac { v^{ 2 }(M+m) }{ gm }$

In collisions we cannot say that energy is conserved always. I think that the distance that we get by this method is the limiting case, the person cannot land any further but he may land before completing this distance.