Question

a person throws a rock at 3 M/s down over the edge of a very tall cliff on Earth how far will the rock have fallen in 4 seconds if the rock never hit the bottom?​

Answer 1

The rock will be at 90.4 m from the top of the cliff.

Explanation:

The rock is thrown with the “initial velocity” 3 m/s. We need to find how much distance does the rock traveled in 4 seconds (t).

From the “kinematic equations” take

$s=u t+\frac{1}{2} a t^{2}$

Where, “s” is distance traveled, “u” initial velocity of the object, “t” time the object traveled and “a” acceleration due to gravity is $9.8 \mathrm{m} / \mathrm{s}^{2}.$

Substitute the given values in the above formula,

$s=3 \times 4+\frac{1}{2} \times 9.8 \times 4^{2}$

$s=12+\frac{1}{2} \times 9.8 \times 16$

$s=12+\frac{1}{2} \times 156.8$

$s=12+78.4$

$s=90.4$

The rock is at height of 90.4 m from the top of the cliff.