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

`a =-153.6\ m/s^(2)`

Step-by-step 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=(v_(f)-v_(0))/(t)`

and, computing

`a =(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.