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Motion under gravity basic concept

This topic is very interesting our daily life we see the example and experience and this was deeply observed by Newton and given Newton's Law of gravitation.

Acceleration under gravity when we hold an object in our hand and release that object then object free fall earth attracts this object towards its center of mass during this we observe the object travel different distance in different time interval like in first second less distance in next second more distance than first second and in third second more distance than 2nd second so here time interval is constant suppose 1 second but distance is increasing means speed is increasing every second when any object speed increases whenever force is apply either it may be push or pull see picture below 

                                           

Motion under gravity basic concept

 hence velocity is changing so acceleration must be there if acceleration is there its means force is involved so earth is pulling object with gravitational force towards its center of mass. we know that per second change of velocity is called acceleration here per second increase in velocity is called acceleration we observe this acceleration value in different places on earth then we found this value is same any places on the earth but why ? because earth surface is same everywhere hence its pulling value is same all places latter we will see some places on mountain its value is slightly changes whenever we go down the earth radius its value decreases and also when we go up from the earth surface its value decreases but it remains constant on the surface of earth at sea level everywhere hence this gravitation force acts every object on the earth so we have made a special symbol of this acceleration due to gravity "g" and scientist checked this value comes g = 9.8 m/s² hence in every one second its speed will increase by 9.8 m/s like in first second 9.8 m/s in next second 19.6 m/s in third second 29.4 m/s and so on now acceleration is a vector quantity hence its direction is permanent down towards earth center. important point acceleration does not depends upon mass so whether object is small or big on both cases acceleration will be same that is g it only depend upon distance from earth surface and time suppose two object of mass are respectively  1 kg and 10 kg released from a building top height 60 m free fall motion (air resistance neglected )which one will appear on the ground first simple acceleration does not depend upon mass and height for both the object is same so time taken to fall on the ground must be same for both object because it depends upon time and distance from earth. now this is acceleration hence all kinematics formula will be applicable here but motion must be rectilinear (straight line motion only) when an object free fall under gravity this motion is called straight line or rectilinear motion we will use acceleration here g.

                     +                                         -

                     ↑                                         ↓  

                     ↑                                         ↓

                     ↑                                         ↓

                     ↑                                         ↓

                     ↑                                         ↓

                     ↑                                         ↓                                       

                     ↑                                         ↓                

                      🔴                                                     🔴                

      →→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→→ 

 here for sign of vector we consider upward direction positive and downward direction negative it is arbitrary we can choose opposite also but generally in our Cartesian coordinate system upward direction is taken as positive and downward is taken as negative but remember g direction is always downward because earth always pull towards its center now apply rectilinear motion equation.

upward motion

v = u -gt  (here velocity of object is in upward direction so it is taken as positive and g is in downward direction taken as negative)

h = ut - 1/2gt²  (here h is used as height )

v² = u² -2gh 

downward motion

v = -u -gt

h = -ut - 1/2gt²

v² = u² +2gh  (here g and h are are negative )

now we will see some basic question on this equation.

Q A stone is dropped from a top of a cliff and is found to travel 44.1 m in the last second before it reaches the ground find height of the cliff.

Ans read question carefully and visualize in your mind before solving any numerical question and note down given parameter and the parameter to find then next step find the relation between given parameter and those parameter to be find.

here it is clear that anything drop and start means its initial velocity is zero u = 0 like a ball drop, a car start means u = 0 no body will tell, you have to understand this word  

again in last one second distance travel is 44.1 m given 

let total height of cliff be h .

now we have study the equation distance travel in nth second remember Snth = u +1/2a(2n -1) here given Snth = 44.1 m , u = 0 put and find value of n 

44.1  = 0+1/2*9.8(2n-1) or 44.1*2 = 2n*9.8 - 9.8 0r n = 5 here total time taken is 5 s

now use equation in 5 s distance travel will be height of cliff 

h = -u -1/2gt² = 0 - 1/2*9.8*5*5 h = 122.5 m.

Q A stone is dropped from a baloon at an altitude of 300 m how long the stone will take to reach the ground if baloon is ascending with a velocity of  5 m/s , if descending with velocity of 5 m/s , if stationary 

Ans here read question carefully what is given first 

h = 300 m, a = g = 9.8 m/s² downward 

important point when two body are moving together and whenever we release one body the velocity of release body will be same whatever speed was together then after gravity will acts upon it then here what will be initial velocity of stone same as velocity of baloon in case of ascending upward with 5 m/s now 

h = -300 m ( here displacement is downward )

g  = -9.8 approx g = 10

u = 5 m/s 

then use equation h = ut+1/2at² or -300  = 5t -1/2*10t²

after solving this equation 5t² - 5t -300 =0 or t² -t - 60 = 0 solve this quadratic equation you will get value of t now in second case when baloon is descending what will be u here just u = -5 m/s other parameter same now in third case when baloon is stationary then u = 0 other parameter will be same put value and get value of time .

thanks for reading 

Motion under gravity basic concept

dated 6th May 2018