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)
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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
[tex]\boldsymbol{S=ut+\frac{1}{2} at^{2} }[/tex]
[tex]\boldsymbol{\therefore 20=0(t)+\frac{1}{2}\times (10)\times t^{2} }[/tex]
[tex]\\\\\boldsymbol{\therefore 5t^{2} =20 }\\\\\boldsymbol{\therefore t^{2} =4 }\\\\\boldsymbol{\therefore t =\sqrt{4} }\\\\\boldsymbol{\therefore t =2\:sec}\\\\[/tex]
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
[tex]\boldsymbol{S=ut+\frac{1}{2} at^{2} }[/tex]
[tex]\boldsymbol{\therefore 20=0(t)+\frac{1}{2}\times (10)\times t^{2} }[/tex]
[tex]\\\\\boldsymbol{\therefore 5t^{2} =20 }\\\\\boldsymbol{\therefore t^{2} =4 }\\\\\boldsymbol{\therefore t =\sqrt{4} }\\\\\boldsymbol{\therefore t =2\:sec}\\\\[/tex]
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
[tex]\boldsymbol{\therefore v=0+(g)(t)}\\\\\boldsymbol{\therefore v=(10)\times(2)}\\\\\boldsymbol{\therefore v=20\:ms^{-1} }[/tex]
Therefore the stone Q hits the ground with a velocity of 20 ms⁻¹.
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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 [tex]S= ut +\frac{1}{2} at^{2}[/tex]
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: [tex]S=ut +\frac{1}{2} at^{2}[/tex]
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: [tex]v= u+ at[/tex]
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.
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