Question

A hailstone traveling with a velocity of 43 meters/second comes to a virtual stop 0.28 seconds after hitting water. What is the magnitude of its acceleration in the water?

Answer 1

\( a = \dfrac{\Delta v}{\Delta t} = \dfrac{43-0}{0.28- 0}
= \dfrac{43}{0.28} = 153.57 ~~ ( 2~ sf)\)