A man found 90 N weight of an object of 9 kg mass at a place. From the same place the object was dro

Here,

Given data are:

Weight of the object (W) = 90 N

Mass of the object (m) = 9 kg

Time taken to fall (t) = 2 s

1.Ans:

To solve the problem of the object dropped from a certain height, we can use the equations of motion under gravity.

The weight W is related to mass m and gravitational acceleration g by the equation:

$\rm W$= $\rm m$ ⋅ $\rm g$

From this we can find value of g:

$\rm g$= $\frac{W}{m}$

$\rm or,$ $\rm g$= $\frac{90}{9}$

$\rm g$= $\rm 10 ms^{-2} $

The distance h fallen in time t under constant acceleration is given by:

(here original formula $\rm h$= $\rm ut $ + $\frac{1}{2}$$\rm gt^{2} $ is not used. It is because the term of ut is neglected as initial velocity(u) is 0 when it is dropped)

$\rm h$= $\frac{1}{2}$$\rm gt^{2} $

$\rm h$= $\frac{1}{2}$$\rm (10)(2)^{2} $

$\rm h$= $\rm 20 m$

Hence, the object was dropped from the height of 20 m.

2.Ans:

For a stone of 500 g, h is same i.e. 20m and g's valuse is same as for the body in first case $\rm 10 ms^{-2} $

Finding time taken by the stone of 500 g :

$\rm h$= $\frac{1}{2}$$\rm gt^{2} $

$\rm t^2$ = $\frac{2h}{g}$

$\rm t^2$ = $\frac{2.20}{10}$

$\rm t^2$ = $\rm 4$

$\rm t$= $\sqrt{4}$

$\rm t$= $\rm 2 s$

The object was dropped from a height of 20 meters.

A stone of mass 500 g dropped from this height will take 2 seconds to reach the ground, which is the same time as the initial object due to gravity acting equally on all masses in free fall (assuming no air resistance). It is because acceleration due to gravity is same for all bodies irrespective of their masses.