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

Answer:

The rock will fall 90.4m downwards

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

$s = ut + \frac{1}{2}at^{2}$

$v^{2} = u^{2} +2as$

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

$s = ut + \frac{1}{2}at^{2}$

Substituting the values in the equation gives,

$s = (-3)x4 + \frac{1}{2}(-9.8)4^{2}$

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.