Aperson fires a 38 gram bullet straight up into the air. it rises, then falls straight back down, striking the ground with a speed of 345 m/s. the bullet embeds itself into the ground a distance of 8.9 cm before coming to a stop. what force does the ground exert on the bullet? express your answer in newtons.

Force exerted = 25.41 kN

Explanation:

We have equation of motion

      v² = u²+2as

u = 345 m/s, s = 8.9 cm = 0.089 m, v = 0 m/s

     0² = 345²+2 x a x 0.089

       a = -668679.78 m/s²

Force exerted = Mass x Acceleration

Mass of bullet = 38 g = 0.038 kg

Acceleration = 668679.78 m/s²

Force exerted = 25409.83 N = 25.41 kN

i think the answer is a, hope this is right, that would be my best guess.