Answer to Question #153540 in Physics for joy

Question

If the stone in fig A enters the water at 41.6 ms^-1, and takes 0.6s to travel the 3 meters to the bottom of the pond, what is its average acceleration in the pond water?

Answer 1

The distance travelled under a constant acceleration is given as follows:

$$d = v_0t -\dfrac{at^2}{2}$$

where $d = 3m$ is the depth of the pond, $t = 0.6s$ is time of movement, $v_0 = 41.6m/s$ is the initial velocity, and $a$ is the required average acceleration. Expressing $a$, find:

$$a = \dfrac{2v_0}{t} - \dfrac{2d}{t^2}\\ a = \dfrac{2\cdot 41.6}{0.6} - \dfrac{2\cdot 3}{0.6^2} = 122m/s^2$$

Answer. 122 m/s^2.