Question

Rahman drops two stones P and Q each of mass 1 kg and 2 kg simultaneously from

the top of a tower of height 20 m. Both start from rest and fall freely downwards. Find

the time taken by the stone P to reach the ground. Does the stone Q reach the ground at

the same time? Find the velocity of the stone Q with which it hits the ground.

(Take g = 10 m/s2)​

Answer 1

The time taken by the stone P to reach the ground is 2 seconds. The stone P as well as Q, both reach the ground at the same time. The stone Q hits the ground with a velocity of 20 ms⁻¹.

Given:

The mass of the stone P = 1 kg

The mass of the stone Q = 2 kg

The height of the tower = 20 m

The acceleration due to gravity (g) = 10 ms⁻²

To Find:

(i) The time taken by the stone P to reach the ground?

(ii) The time taken by the stone Q to reach the ground?

(iii) The velocity with which the stone Q hits the ground?

Solution:

The mass of the stone P = 1 kg

The mass of the stone Q = 2 kg

The initial velocity of the stone P = 0

The initial velocity of the stone P = 0

→ The acceleration due to gravity (g) = 10 ms⁻²

∴ The acceleration of the stone P = g = 10 ms⁻² (downwards)

∴ The acceleration of the stone Q = g = 10 ms⁻² (downwards)

(i) Calculating the time taken by the stone P to reach the ground:

→ Using the Second Equation of motion: S = ut + (1/2)at²

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

                               $\boldsymbol{\therefore 20=0(t)+\frac{1}{2}\times (10)\times t^{2} }$

                                      $\\\\\boldsymbol{\therefore 5t^{2} =20 }\\\\\boldsymbol{\therefore t^{2} =4 }\\\\\boldsymbol{\therefore t =\sqrt{4} }\\\\\boldsymbol{\therefore t =2\:sec}$

Therefore the time taken by the stone P to reach the ground is 2 seconds.

(ii) Calculating the time taken by the stone Q to reach the ground:

→ Using the Second Equation of motion: S = ut + (1/2)at²

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

                               $\boldsymbol{\therefore 20=0(t)+\frac{1}{2}\times (10)\times t^{2} }$

                                      $\\\\\boldsymbol{\therefore 5t^{2} =20 }\\\\\boldsymbol{\therefore t^{2} =4 }\\\\\boldsymbol{\therefore t =\sqrt{4} }\\\\\boldsymbol{\therefore t =2\:sec}$

Therefore the time taken by the stone Q to reach the ground is 2 seconds.

→ Hence the stone P as well as Q, both reach the ground at the same time.

(iii) Calculating the velocity with which the stone Q hits the ground:

→ Using the First Equation of motion: v = u + at

                                     $\boldsymbol{\therefore v=0+(g)(t)}\\\\\boldsymbol{\therefore v=(10)\times(2)}\\\\\boldsymbol{\therefore v=20\:ms^{-1} }$

Therefore the stone Q hits the ground with a velocity of 20 ms⁻¹.

#SPJ1

Answer 2

Answer:

The time taken by the stone P to reach the ground is 2 seconds. The stone P as well as Q, both reach the ground at the same time. The stone Q hits the ground with a velocity of 20 ms⁻¹.

Explanation:

We are given some of the information that is-\ mass of the stone P = 1 kg, The mass of the stone Q = 2 kg, The height of the tower = 20 m, The acceleration due to gravity (g) = 10 ms⁻²

We have to find- t he time taken by the stone P to reach the ground? The time taken by the stone Q to reach the ground? The velocity with which the stone Q hits the ground?

The initial velocity of the stone P = 0

The initial velocity of the stone P = 0

The acceleration due to gravity (g) is 10 ms⁻²

Therefore he acceleration of the stone P = g = 10 ms⁻² (in downwards directions)

The acceleration of the stone Q = g = 10 ms⁻² (in downwards directions)

(i) Calculating the time taken by the stone P to reach the ground:

Using the Second Equation of motion $S= ut +\frac{1}{2} at^{2}$

After substituting all the value we will the value of t that is 2 m per second.

Therefore the time taken by the stone P to reach the ground is 2 seconds.

(ii) Calculating the time taken by the stone Q to reach the ground:

Using the Second Equation of motion: $S=ut +\frac{1}{2} at^{2}$

After substituting all the values we get t is equal to 2 m per second

Therefore the time taken by the stone Q to reach the ground is 2 seconds.

Hence the stone P as well as Q, both reach the ground at the same time.

(iii) Calculating the velocity with which the stone Q hits the ground:

Using the First Equation of motion: $v= u+ at$

After substituting the values we get v is equal to 20 m per sec.

Therefore the stone Q hits the ground with a velocity of 20 ms⁻¹.

#SPJ1

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