### Chapter : Motion in one Dimension

__Motion Under Gravity __

__Motion Under Gravity__

**Ideal Motion :**

The most important example of motion in a straight line with constant acceleration is motion under gravity. In case of motion under gravity unless stated it is taken for granted that.

**(i)**The acceleration is constant, i.e. = = 9.8 m/s

^{2}and directed vertically downwards.

**(ii)**The motion is in vacuum i.e. viscous force or thrust of the medium has no effect on the motion.

Now in the light of above assumptions, there are two possibilities.

**10.1 Body Falling Freely Under Gravity :**

Taking initial position as origin and direction of motion (i.e. downward direction) as positive, here we have u = 0 (as body starts from rest)

a = + g

(as acceleration is in the direction of motion)

So, if the body acquires velocity v after falling a distance h in time t, equations of motion viz v = u + at , s = ut + at

^{2}and v

^{2}= u

^{2}+ 2as reduces to

v = gt ....(1)

h = gt

^{2}....(2)

v

^{2}= 2gh ....(3)

These equations can be used to solve most of the problems of freely falling as

If t is given | If h is given | If v is given |
---|---|---|

From eq. (1) & (2) | From eq. (2) & (3) | From eq. (3) & (1) |

v = gt | t= | t= |

h = gt^{2} | v= | |

**Note :**

**(a)**If the body is dropped from a height H, as in time t, it has fallen a distance h from its initial position, the height of the body from the ground will be h' = H – h, with h = 1/2 gt

^{2}.

**(b)**As h = (1/2) gt

^{2}i.e. h ∝ t

^{2}, distance fallen in time t, 2t, 3t etc. will be in the ratio of 1

^{2}:

^{2}2: 3

^{2}: ––––––– i.e. square of integers.

**(c)**The distance fallen in n

^{th}sec.,

h

_{n}– h

_{n –1 }= (1/2) g(n)

^{2}– (1/2) g(n – 1)

^{2}= (1/2) g(2n –1)

So distance fallen in I

^{st}, II

^{nd}, III

^{rd}sec will be in the ratio 1 : 3 : 5 i.e. odd integers only.

**10.2 Body is projected vertically up :**

Taking initial position as origin and direction of motion (i.e. vertically up) as positive, here we have v = 0 [as at the highest point, velocity = 0], a = – g [as acceleration is downwards while motion upwards].

So, if the body is projected with velocity u and reaches the highest point at a distance h above the ground in time t, the equations of motion viz v = u + at, s = ut + at

^{2}and v

^{2}= u

^{2}+ 2as reduces to

0 = u – gt

h = ut – 1/2 gt

^{2}

and 0 = u

^{2}– 2gh

or u = gt ....(1)

h = 1/2 gt

^{2}...(2)

(∴ u = gt), u

^{2}= 2 gh ....(3)

These equations can be used to solve most of the problems of bodies projected vertically up as, if

If t is given | If h is given | If v is given |
---|---|---|

From eq. (1) & (2) | From eq. (2) & (3) | From eq. (3) & (1) |

u = gt | t= | t= u/g |

h = gt^{2} | u = | h = u^{2}/2g |

**Discussion :**

**From cases (10.1) and (10.2) it is clear that :**

**(a)**In case of motion under gravity for a given body, mass, acceleration and mechanical energy remains constant while speed, velocity, momentum, kinetic energy and potential energy changes.

**(b)**The motion is independent to the mass of the body as in any equation of motion mass is not involved. This is why a heavy and light body when released from same height reaches the ground simultaneously and with same velocity.

i.e. t =

and v =

However, momentum, kinetic energy or potential energy depends on the mass of the body (all ∝ mass)

**(c)**As from case (b) time taken to reach a height h,

t

_{U}=

And from case (a) time taken to fall down through a distance h,

t

_{D}=

so t

_{U}= t

_{D}=

So in case of motion under gravity time taken to go up is equal to the time taken to fall down through the same distance.

**(d)**If a body projected vertically up reaches a height h then from case (b), u = and if a body falls freely through a height h from case (a),

v =

So 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

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