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Acceleration produced in a body by the force of gravity, is called acceleration due to gravity. It is represented by the symbol \(g\). Value of \(g\) is \(9.8m/{s^2}\) or, \(980cm/{s^2}\) or, \(32ft/{s^2}\).
In the absence of air, it is found that all bodies (irrespective of the size, weight or composition) fall with same acceleration near the surface of the earth. The motion of a body falling towards the earth from a small altitude (\(h < < R\)) is called motion under gravity. Free fall means acceleration of body is equal to acceleration due to gravity.
Here Five Case may be arises.
(A) ▉ Body is Projected Vertically Upward:
🔷🔷🔷🔷🔷Equation of Motion:
Taking initial position as origin and direction of motion as positive,
\(a = - g\), As acceleration is downwards while motion is upward.
So, If the body is projected with initial velocity \(u\) and after time \(t\) it reaches up to height \(h\) then,
(1) \(v = u - gt\)
(2) \(h = ut - \frac{1}{2}g{t^2}\)
(3) \({v^2} = {u^2} - 2gh\)
(4) \({h_n} = u - \frac{1}{2}g\left( {2n - 1} \right)\)
🔷🔷🔷 Equation of Motion Becomes:
For maximum height, \(v = 0\), So from above equation becomes
(1) \(u = gt\) ... ... ... (i)
(2) \(h = ut - \frac{1}{2}g{t^2} = \left( {gt} \right)t - \frac{1}{2}g{t^2} = g{t^2} - \frac{1}{2}g{t^2} = \frac{1}{2}g{t^2}\)= \(\frac{1}{2}g{\left( {\frac{u}{g}} \right)^2} = \frac{1}{2}\frac{{{u^2}}}{g}\) ... ... ... (ii)
(3) \({u^2} = 2gh\) ... ... ... (iii)
(4) \({h_n} = u - \frac{1}{2}g\left( {2n - 1} \right)\) ... ... ... (iv)
🔴 Time of Ascent: From equation (i), Time of ascent \({t_1} = \frac{u}{g}\) Therefore, Time of ascent = \({t_1} = \frac{u}{g}\)
🔴 Time of descent:
\(\frac{1}{2}\frac{{{u^2}}}{g} = \frac{1}{2}gt_2^2\)
or, \(t_2^2 = \frac{{{u^2}}}{{{g^2}}}\)
or, \({t_2} = \frac{u}{g}\)
When the body comes down toward ground, its velocity is \(0\) and height from the ground is \(\frac{1}{2}\frac{{{u^2}}}{g}\). Its acceleration is taken positive toward ground. If time of descent is \({t_2}\) then,
Therefore, Time of descent = \({t_2} = \frac{u}{g}\) In case of motion under gravity, time taken to go up is equal to the time taken to fall down through the same distance. So,
🔴 Total Time of Flight: Time of ascent (\({t_1}\)) = Time of descent (\({t_2}\)) = \(\frac{u}{g}\)
Therefore, the Total time of flight \(T = {t_1} + {t_2} = \frac{u}{g} + \frac{u}{g} = \frac{{2u}}{g}\)
🔴 Maximum Height Reach: \(h = ut - \frac{1}{2}g{t^2} = \left( {gt} \right)t - \frac{1}{2}g{t^2} = g{t^2} - \frac{1}{2}g{t^2} = \frac{1}{2}g{t^2}\)= \(\frac{1}{2}g{\left( {\frac{u}{g}} \right)^2} = \frac{1}{2}\frac{{{u^2}}}{g}\) ... ... ... (ii)
Therefore, Maximum height = \(h = \frac{1}{2}g{t^2} = \frac{1}{2}\frac{{{u^2}}}{g}\) And, \({u^2} = 2gh\)
or, Maximum height reached \(h = \frac{{{u^2}}}{{2g}}\)
🔴 Velocity on Reaching the ground:
\({v^2} = 2gh = 2g \times \frac{1}{2}\frac{{{u^2}}}{g} = {u^2}\)
or, \(v = u\)
Thus in case of motion under gravity, the speed with which a body is projected up is equal to the speed with which it comes back to the point of projection.
Knowledge Points:
🔯 It is clear that both quantities do not depend upon the mass of the body or we can say that in absence of air resistance, all bodies fall on the surface of the earth with the same rate.
🔯 In case of motion under gravity for a given body, mass, acceleration and mechanical energy remain constant while speed, velocity, momentum, kinetic energy and potential energy change.
🔯 The motion is independent of the mass of the body, as in any equation of motion, mass is not involved. That is why a heavy and light body when released from the same height, reach the ground simultaneously and with same velocity i.e., \(t = \frac{u}{g}\) or \(\sqrt {\frac{{2h}}{g}} \) and \(v = u = \sqrt {2gh} \).
🔯 In case of motion under gravity time taken to go up is equal to the time taken to fall down through the same distance. Time of ascent (\({t_1}\))= Time of decent = \({t_2}\) = \(\frac{u}{g}\) And total time of flight = \(t = {t_1} + {t_2} = \frac{{2u}}{g}\)
🔯 (C) ▉ Body is Projected Vertically Downward from Some height (Initial Velocity is Zero):
Equation of Motion:
Taking initial position as origin and direction of motion (downward direction) as a positive, here we have,
\(u = 0\) As the body start from rest.
\(a = + g\) As Acceleration is in the direction of motion.
So, If the body dropped some height and after certain a time, it reaches the ground, then,
(1) \(v = gt\) ... ... ... (i)
(2) \(h = \frac{1}{2}g{t^2}\) ... ... ... (ii)
(3) \({v^2} = 2gh\) ... ... ... (iii)
(4) \({h_n} = \frac{1}{2}g\left( {2n - 1} \right)\) ... ... ... (iv)
(D) ▉ Body is Projected vertically downward with some initial Speed:
Equation of Motion
Taking initial position as origin and direction of motion (downward direction) as a positive, we have,
(1) \(v = u + gt\) ... ... ... (i)
(2) \(h = ut + \frac{1}{2}g{t^2}\) ... ... ... (ii)
(3) \({v^2} = {u^2} + 2gh\) ... ... ... (iv)
(4) \({h_n} = u + \frac{1}{2}g\left( {2n - 1} \right)\) ... ... ... (iv)
(E) ▉ Body is Projected vertically upward with some initial speed from a certain height:
Equation of Motion:
Taking initial position as origin and direction of motion (upward direction) as negative, here we have,
Initial velocity = \( + u\)
\(a = - g\) As Acceleration is downward while motion is upward.
(1) \(v = u - gt\) ... ... ... (i)
(2) \(h = ut - \frac{1}{2}g{t^2}\) ... ... ... (ii)
(3) \({v^2} = {u^2} - 2gh\)
(4) \({h_n} = u - \frac{1}{2}g\left( {2n - 1} \right)\)
Maximum height attained by the body:
For maximum height \(v = 0\)
So from above equation,
\(u = gt\), So, \(t = \frac{u}{g}\)
\(h = \frac{1}{2}g{t^2}\) or, \(h = \frac{1}{2}g{t^2} = \frac{1}{2}g{\left( {\frac{u}{g}} \right)^2} = \frac{1}{2}\frac{{{u^2}}}{g}\)
\({u^2} = 2gh\)
Total Height attained by the body = \({H_{\max }} = H + h = H + \frac{1}{2}\frac{{{u^2}}}{g}\)
Distance travelled by the body = \(H + 2h = H + \left( {2 \times \frac{1}{2}\frac{{{u^2}}}{g}} \right) = H + \frac{{{u^2}}}{g}\)
Time taken to reach the ground:
\(H = ut - \frac{1}{2}g{t^2}\)
or, \(\frac{1}{2}g{t^2} - ut - H = 0\)
or, \(g{t^2} - 2ut - 2H = 0\)
After solving this equation we get the result. <! paragraph Style End --->
In the absence of air, it is found that all bodies (irrespective of the size, weight or composition) fall with same acceleration near the surface of the earth. The motion of a body falling towards the earth from a small altitude (\(h < < R\)) is called motion under gravity. Free fall means acceleration of body is equal to acceleration due to gravity.
Here Five Case may be arises.
(A) ▉ Body is Projected Vertically Upward:
🔷🔷🔷🔷🔷Equation of Motion:
Taking initial position as origin and direction of motion as positive,
\(a = - g\), As acceleration is downwards while motion is upward.
So, If the body is projected with initial velocity \(u\) and after time \(t\) it reaches up to height \(h\) then,
(1) \(v = u - gt\)
(2) \(h = ut - \frac{1}{2}g{t^2}\)
(3) \({v^2} = {u^2} - 2gh\)
(4) \({h_n} = u - \frac{1}{2}g\left( {2n - 1} \right)\)
🔷🔷🔷 Equation of Motion Becomes:
For maximum height, \(v = 0\), So from above equation becomes
(1) \(u = gt\) ... ... ... (i)
(2) \(h = ut - \frac{1}{2}g{t^2} = \left( {gt} \right)t - \frac{1}{2}g{t^2} = g{t^2} - \frac{1}{2}g{t^2} = \frac{1}{2}g{t^2}\)= \(\frac{1}{2}g{\left( {\frac{u}{g}} \right)^2} = \frac{1}{2}\frac{{{u^2}}}{g}\) ... ... ... (ii)
(3) \({u^2} = 2gh\) ... ... ... (iii)
(4) \({h_n} = u - \frac{1}{2}g\left( {2n - 1} \right)\) ... ... ... (iv)
🔴 Time of Ascent: From equation (i), Time of ascent \({t_1} = \frac{u}{g}\) Therefore, Time of ascent = \({t_1} = \frac{u}{g}\)
🔴 Time of descent:
\(\frac{1}{2}\frac{{{u^2}}}{g} = \frac{1}{2}gt_2^2\)
or, \(t_2^2 = \frac{{{u^2}}}{{{g^2}}}\)
or, \({t_2} = \frac{u}{g}\)
When the body comes down toward ground, its velocity is \(0\) and height from the ground is \(\frac{1}{2}\frac{{{u^2}}}{g}\). Its acceleration is taken positive toward ground. If time of descent is \({t_2}\) then,
Therefore, Time of descent = \({t_2} = \frac{u}{g}\) In case of motion under gravity, time taken to go up is equal to the time taken to fall down through the same distance. So,
🔴 Total Time of Flight: Time of ascent (\({t_1}\)) = Time of descent (\({t_2}\)) = \(\frac{u}{g}\)
Therefore, the Total time of flight \(T = {t_1} + {t_2} = \frac{u}{g} + \frac{u}{g} = \frac{{2u}}{g}\)
🔴 Maximum Height Reach: \(h = ut - \frac{1}{2}g{t^2} = \left( {gt} \right)t - \frac{1}{2}g{t^2} = g{t^2} - \frac{1}{2}g{t^2} = \frac{1}{2}g{t^2}\)= \(\frac{1}{2}g{\left( {\frac{u}{g}} \right)^2} = \frac{1}{2}\frac{{{u^2}}}{g}\) ... ... ... (ii)
Therefore, Maximum height = \(h = \frac{1}{2}g{t^2} = \frac{1}{2}\frac{{{u^2}}}{g}\) And, \({u^2} = 2gh\)
or, Maximum height reached \(h = \frac{{{u^2}}}{{2g}}\)
🔴 Velocity on Reaching the ground:
\({v^2} = 2gh = 2g \times \frac{1}{2}\frac{{{u^2}}}{g} = {u^2}\)
or, \(v = u\)
Thus in case of motion under gravity, the speed with which a body is projected up is equal to the speed with which it comes back to the point of projection.
Knowledge Points:
🔯 It is clear that both quantities do not depend upon the mass of the body or we can say that in absence of air resistance, all bodies fall on the surface of the earth with the same rate.
🔯 In case of motion under gravity for a given body, mass, acceleration and mechanical energy remain constant while speed, velocity, momentum, kinetic energy and potential energy change.
🔯 The motion is independent of the mass of the body, as in any equation of motion, mass is not involved. That is why a heavy and light body when released from the same height, reach the ground simultaneously and with same velocity i.e., \(t = \frac{u}{g}\) or \(\sqrt {\frac{{2h}}{g}} \) and \(v = u = \sqrt {2gh} \).
🔯 In case of motion under gravity time taken to go up is equal to the time taken to fall down through the same distance. Time of ascent (\({t_1}\))= Time of decent = \({t_2}\) = \(\frac{u}{g}\) And total time of flight = \(t = {t_1} + {t_2} = \frac{{2u}}{g}\)
🔯 (C) ▉ Body is Projected Vertically Downward from Some height (Initial Velocity is Zero):
Equation of Motion:
Taking initial position as origin and direction of motion (downward direction) as a positive, here we have,
\(u = 0\) As the body start from rest.
\(a = + g\) As Acceleration is in the direction of motion.
So, If the body dropped some height and after certain a time, it reaches the ground, then,
(1) \(v = gt\) ... ... ... (i)
(2) \(h = \frac{1}{2}g{t^2}\) ... ... ... (ii)
(3) \({v^2} = 2gh\) ... ... ... (iii)
(4) \({h_n} = \frac{1}{2}g\left( {2n - 1} \right)\) ... ... ... (iv)
(D) ▉ Body is Projected vertically downward with some initial Speed:
Equation of Motion
Taking initial position as origin and direction of motion (downward direction) as a positive, we have,
(1) \(v = u + gt\) ... ... ... (i)
(2) \(h = ut + \frac{1}{2}g{t^2}\) ... ... ... (ii)
(3) \({v^2} = {u^2} + 2gh\) ... ... ... (iv)
(4) \({h_n} = u + \frac{1}{2}g\left( {2n - 1} \right)\) ... ... ... (iv)
(E) ▉ Body is Projected vertically upward with some initial speed from a certain height:
Equation of Motion:
Taking initial position as origin and direction of motion (upward direction) as negative, here we have,
Initial velocity = \( + u\)
\(a = - g\) As Acceleration is downward while motion is upward.
(1) \(v = u - gt\) ... ... ... (i)
(2) \(h = ut - \frac{1}{2}g{t^2}\) ... ... ... (ii)
(3) \({v^2} = {u^2} - 2gh\)
(4) \({h_n} = u - \frac{1}{2}g\left( {2n - 1} \right)\)
Maximum height attained by the body:
For maximum height \(v = 0\)
So from above equation,
\(u = gt\), So, \(t = \frac{u}{g}\)
\(h = \frac{1}{2}g{t^2}\) or, \(h = \frac{1}{2}g{t^2} = \frac{1}{2}g{\left( {\frac{u}{g}} \right)^2} = \frac{1}{2}\frac{{{u^2}}}{g}\)
\({u^2} = 2gh\)
Total Height attained by the body = \({H_{\max }} = H + h = H + \frac{1}{2}\frac{{{u^2}}}{g}\)
Distance travelled by the body = \(H + 2h = H + \left( {2 \times \frac{1}{2}\frac{{{u^2}}}{g}} \right) = H + \frac{{{u^2}}}{g}\)
Time taken to reach the ground:
\(H = ut - \frac{1}{2}g{t^2}\)
or, \(\frac{1}{2}g{t^2} - ut - H = 0\)
or, \(g{t^2} - 2ut - 2H = 0\)
After solving this equation we get the result. <! paragraph Style End --->
(h)
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