Answer to Question #153465 in Physics for joy

Question

(1). The bird drops the stone from a height of 88 m above the water surface. Initially, the stone has zero vertical velocity. How long will it take the stone to reach the surface of the pond?(Assume air resistance is negligible.) 

(2). If the stone in bird 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

(1) $h=gt^2/2\to t=\sqrt{\frac{2h}{g}}= \sqrt{\frac{2\cdot88}{9.8}}=4.24(s)$

(2) $h=v_0t-at^2/2\to a=2(v_0t-h)/t^2=2(41.6\cdot0.6-3)/0.6^2=$

$=122(m/s^2)$