Speed of the arrow just before it will hit the ground can be calculated by energy conservation
[tex]\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f[/tex]
[tex]\frac{1}{2}(50)^2 + 9.8(10) = \frac{1}{2}v_f^2 + 0[/tex]
by solving above equation
[tex]v_f = 51.9 m/s[/tex]
now this will lodge itself 15 cm into the ground
so after lodging itself to 15 cm the speed of arrow becomes zero
now we will have
[tex]v_f^2 - v_i^2 = 2a d[/tex]
[tex]0 - 51.9^2 = 2(a)(0.15)[/tex]
[tex]a = -8986.7 m/s^2[/tex]
Now by Newton's II law net force is given as
[tex]F = ma[/tex]
[tex]F = 0.050 \times 8986.7[/tex]
[tex]F = 449.3 N[/tex]
