Tuesday, 27 March 2018

Projectile motion equations and formula

  Projectile motion equations and formula

Projectile motion is our topic for today this topic is easy but conceptual and best example of 2-D motion very first what is projectile motion equation .It is define as any body or particle having m mass negligible  size if we through vertically upward then body moves in straight line similarly if we through body in horizontal line again it will move in horizontal straight line so both motion are straight line but when we through body between horizontal and vertical plan making some angle with horizontal line then body moves in a parabolic shape this is called projectile motion and path travel by projectile is called projectile motion equation see picture below.

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                                            Projectile motion equation

This motion is also called free fall motion because when we project body only gravitational force acts upon the body no any other force acts on body hence called free fall motion so all fundamental concept is applicable for projectile motion.
important point this is plan motion 2-D motion but we analyse this motion by separate two straight line motion one along x axis and other along y axis why ? because we have no equation of motion developed for plan 2-D motion now till so we use two separate straight line motion for solving equation of projectile motion other important point is velocity along x axis is ucos𝛳 and velocity along y axis is usin𝛳 now you tell me which velocity component is responsible for projectile to achieve height and in air obviously usin𝛳 is responsible because this is only velocity due to which projectile travel maximum height and this velocity is also variable due to gravity acceleration along y axis acting vertically  downward initially velocity decrease and at maximum height its become zero the projectile start falling down but note at maximum height total velocity will not be zero velocity along x axis will be present there now velocity along x axis is throughout same each and every point is same but why because there is no any acceleration along x axis means acceleration along x axis is zero hence velocity along x axis will be constant so using this all component velocity and acceleration we will find out projectile parameter by trick.

initial velocity along x axis  = ucos𝛳
initial velocity along y axis  = usin𝛳 
acceleration along y axis  = -g
acceleration along x axis  = 0
where u is velocity of projectile and  ð›³ is angle of projection with x axis now we have stated above using straight line motion in two separate axis x and y we will find all projectile parameter you can refer straight line motion .
here are three equation 
v = u+at, s = ut +1/2at²  v²  = u²  +2as
now for projectile motion we will use this equation to find its parameter 
Time of flight this is define as total time taken by projectile in air now we want to find time of flight apply trick we know that velocity along y axis is responsible to kept in air hence use velocity along y axis to find time of flight
we know that initial velocity along y axis  =  usin𝛳 and velocity along y axis at maximum height will be zero hence use equation 
v = u +at  where a = -g
0  = usin𝛳 -gt after simplification  t =  usin𝛳/g this is also call time of accent time taken to travel maximum height and same time taken to decent from maximum height to ground hence total time in air will be T = time of accent +time of decent  =   usin𝛳/g + usin𝛳/g = 2usin𝛳/g so T =      2usin𝛳/g
Alternate method you can use 2nd equation  s = ut +1/2at² here you can use concept displacement along y axis is zero after reaching projectile on the ground this will give total time projectile in air so put all value  0 =  usin𝛳t - 1/2gt² = t(usin𝛳-1/2gt)  = 0
now t = 0 and t  =    2usin𝛳/g here t = 0 gives when projectile was start throwing and t = 2usin𝛳/g  gives when projectile reaches at ground so    t = 2usin𝛳/g is time of flight. 
Maximum height   we know that for maximum height velocity along y axis is responsible hence use third equation    = u²  +2as 
now we know that at maximum height velocity along y axis will be zero so  0 =  (usin𝛳)²  -2gh after simplification  H =  (usin𝛳)²/2g this is maximum height.
Range    this is define as horizontal distance travel along x axis by projectile for this we know that velocity along x axis is responsible to travel distance along x axis and acceleration along x axis is zero hence again use 2nd equation  s = ut +1/2at² 

R  =   ucos𝛳*time of flight  - 1/2*0 =  ucos𝛳* 2usin𝛳/g = u²2 sin𝛳cos𝛳/g  = u²sin2𝛳/g  hence  R  =  u²sin2𝛳/g now for maximum range    sin2𝛳 = 1 so  ð›³ = 45⁰    
Q1 sometime asked question for which two value of ð›³ Range will be same for this if one angle is ð›³ then other angle will be (90-𝛳) for same range  
Q2 Find the speed of projectile after time t.
for this question how you can approach after time t we have to find speed we know that projectile having velocity in both component along x axis and along y axis we also know that velocity along x axis remain constant and is equal to  ucos𝛳
velocity along y axis is variable depend upon time hence can be written as  v  = u+at  =  usin𝛳 -gt hence total velocity will be 
V   =  ucos𝛳i^ + ( usin𝛳 -gt )j^  now for speed take modulus 
|v|  = √(ucos𝛳)² +usin𝛳 -gt )² this is speed of projectile after t time.
Q3 A particle is projected from a tower height 20m from ground with a velocity of 20m/s at an angle of 30⁰ with horizontal find  maximum height from ground, total time of flight,find range, find final speed of hitting ground.   

for any projectile question first you have to take separate two motion along x axis and y axis for this question we already know maximum height formula H = (usin𝛳)²/2g  put the value 
H = (20*sin30)²/2g  = (20*1/2)²/2*10 = 100/20 = 5m
hence maximum height from ground = tower height + maximum height  = 20 + 5  = 25m
for total time of flight you have to approach by trick velocity along y axis is responsible for projectile to be in air or time of flight first you calculate time taken for maximum height use equation 
v  = u +at  final velocity along y will be zero hence t  = usin𝛳/g
t = 20*sin30/10 = 20*1/2*1/10  = 1s 
now after maximum height projectile will fall down and travel 25m distance vertical and initial velocity is zero hence use equation 
s = ut +1/2at²  put all value  = 25  = 0+1/2*10t² after simplification t  = √5  so for time of flight total time = 1+√5 .
now for range we know that velocity along x axis is responsible for range hence use same equation s = ut +1/2at² for range here a =0
R = 20*cos30*time of flight  = 20*√3/2*(1+√5) 
now for final speed we know that velocity along x axis will be same hence velocity along x axis = 20*cos30 now velocity along y axis use equation  v = u+at  initial velocity along y axis at maximum height is zero so v = gt = 10*√5  hence speed will be take modulus of total velocity along x and y axis that is 
V   =  10√3i^ +10√5j^  now take modulus of this.

|V|  = √(10√3)² +(10√5)²  m/s will be speed with which projectile will hit ground .
now sometime asked trajectory this is the path of projectile which is parabolic shown above in picture we can find out this equation lets see how write the 2nd equation   s = ut +1/2at² use separately for x axis and y axis for x axis  x  = ucos𝛳t +0  =  ucos𝛳
for y axis  y  = usin𝛳t - 1/2gt² now from x  =  ucos𝛳t so t = x/ucos𝛳
put t =  x/ucos𝛳 in y equation we will get equation of trajectory.
now using this concept you can solve any type of projectile problem i hope you have enjoyed learning  Projectile motion equations and formula thanks for reading and sharing.
dated 14th Jul 2018.

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