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A person 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.

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Blacklash

Answer:

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

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