Question

If a stone dropped into a well reaches the water's surface after 3.0 seconds, how far did the stone drop before hitting the water?

Answer 1

s= 0.5 at^2

0.5 x 9.81 x 3^2 = 44m

Answer 2

Explanation:

It is given that,

The stone reaches the water surface is 3 seconds. Let d is the distance covered by the by the stone before hitting the water. Initial speed of the stone, u = 0

Using second equation of motion to find the distance covered.

$d=ut+\dfrac{1}{2}at^2$

Here, a = g

$d=\dfrac{1}{2}gt^2$

$d=\dfrac{1}{2}\times 9.8\times (3)^2$

d = 44.1 meters

So, the stone will cover a distance of 44.1 meters before hitting the water. Hence, this is the required solution.