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 at fall in 4 seconds if the rock never hit the bottom?​

Answer 1

The rock will fall 90.4m downwards

Step-by-step explanation:

Here, the acceleration is constant and it is equal to acceleration due to gravity. Therefore the question depicts a situation of uniformly accelerated motion in a straight line. So, let us refresh the three equations of uniformly accelerated straight line motion.

v = u + at

`s = ut + (1)/(2)at^(2)`

`v^(2) = u^(2) +2as`

where,

u = initial velocity

v = final velocity

s = displacement

a = acceleration

t = time

Since we are dealing with vectors (velocity, acceleration and displacement), we have to take their directions in to account. So we must adopt a coordinate system according to our convenience. Here, we are taking point of throwing as origin, vertically upward direction as positive y axis and vertically downward direction as negative y axis.

Thus,

u = -3m/s (since it is along negative y axis)

a = g = -9.8m/
`s^(2)` (since it is along negative y axis)

t = 4s

s = ?

The only equation that connects all these quantities is

`s = ut + (1)/(2)at^(2)`

Substituting the values in the equation gives,

`s = (-3)x4 + (1)/(2)(-9.8)4^(2)`

s = -12 - 78.4

s = -90.4m

Thus the rock will fall 90.4m downwards. The negative sign indicates that the displacement is along negative y axis which is downwards.