Question

a bullet moving horizontally to the right with a sped of 500m/s strikes a sandbag and penetrates a distance of 10cm. what is the average acceleration in m/s^2 of the bullet?

Answer 1

$s=\frac{v^2-v_0^2}{2a}\to a=\frac{v^2-v_0^2}{2s}=\frac{0^2-500^2}{2\cdot0.1}=1250000(m/s^2)$