A bullet flying at a speed of 400 m / s hits the sand and penetrates to a depth of 20 cm.

A bullet flying at a speed of 400 m / s hits the sand and penetrates to a depth of 20 cm. What was the speed of a bullet at a depth of 10 cm?

Given: V0 (initial bullet speed) = 400 m / s; S1 (bullet path to full stop) = 20 cm (0.2 m); S2 (considered bullet penetration depth) = 0.1 m; the bullet motion is assumed to be uniformly slowed down.

1) Determine the constant acceleration of the bullet: S2 = (V ^ 2 – V0 ^ 2) / 2a = (0 – V0 ^ 2) / 2a and a = V ^ 2 / (2 * S ^ 2) = 400 ^ 2 / ( 2 * 0.2) = -400000 m / s2.

2) Let’s calculate the speed of the bullet at a depth of 10 cm: S1 = (V1 ^ 2 – V0 ^ 2) / 2a; V1 ^ 2 = S1 * 2a + V0 ^ 2 and V1 = √ (S1 * 2a + V0 ^ 2) = √ (0.1 * 2 * (-400000) + 400 ^ 2) = √80000 = 282.84 m /with.