Select the correct answer. A hailstone traveling with a velocity of 43meters/second comes to a virtual stop 0.28 seconds after hitting water. What is the magnitude of its acceleration in the water? A. 12meters/second^2 B 1.5times 10^2meters/second^2 C. 78.6meters/second^2 D. 6.5times 10^-3meters/second^2

To find the magnitude of the acceleration, we can use the formula:<br /><br />\[ a = \frac{\Delta v}{\Delta t} \]<br /><br />where \( a \) is the acceleration, \( \Delta v \) is the change in velocity, and \( \Delta t \) is the change in time.<br /><br />Given:<br />Initial velocity (\( v_i \)) = 43 meters/second<br />Final velocity (\( v_f \)) = 0 meters/second (since it comes to a virtual stop)<br />Time (\( \Delta t \)) = 0.28 seconds<br /><br />Now, let's calculate the change in velocity (\( \Delta v \)):<br /><br />\[ \Delta v = v_f - v_i = 0 - 43 = -43 \, \text{meters/second} \]<br /><br />Next, we can calculate the magnitude of the acceleration:<br /><br />\[ a = \frac{\Delta v}{\Delta t} = \frac{-43}{0.28} \]<br /><br />\[ a = -153.57 \, \text{meters/second}^2 \]<br /><br />Since we are looking for the magnitude of the acceleration, we take the absolute value:<br /><br />\[ |a| = 153.57 \, \text{meters/second}^2 \]<br /><br />Therefore, the correct answer is:<br /><br />B. \( 1.5 \times 10^2 \, \text{meters/second}^2 \)