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

Acceleration=(change in speed)/(time for the change). 43/0.28 = 153.6 m/s^2.

Answer 2

Answer:

$a =-153.6\ m/s^{2}$

Explanation:

Since we are assuming that is acceleration in water is constant, we are going to use one of the uniform accelerated motion formulas. Since we know the initial velocity ($v_{0}=43 m/s$), the final velocity ($v_f=0$) and, the time it takes to stop ($t=0.28s$) so, we are going to use:

$v_{f}=v_{0}+at$,

where a is the acceleration.

Solving for the acceleration

$a=\dfrac{v_{f}-v_{0}}{t}$

and, computing

$a =\dfrac{0-43}{0.28}\\\\a=-153.6\ m/s^{2}$.

Notice that the acceleration is negative. This is because the speed decreases when the hailstone enters the water.