What is projectile motion formula with an example ?
However important about projectile motion. This is plan motion 2-D. But you analyse this motion, by separate two straight line motion. Hence one along x axis, and other along y axis. Do you know why ? .
Because we have no any equation of motion . Which is available for plan 2-D motion till date . So you use two separate straight line motion. For solving equation of projectile motion. And other important point is velocity, along x axis which is ucosα.
Similarly velocity along y axis is usinα. Now you tell me, which velocity component is responsible, for projectile to achieve height. Well in air obviously usinα is responsible. Because this is the only velocity, due to which projectile travel maximum height.
Hence this velocity is also variable, acceleration due to gravity along y axis. Which is acting vertically downward. So initially velocity decrease, and at maximum height, its become zero. Then the projectile starts falling down.
What is plane motion explain its importance in horizontal projectile motion ?
But note at maximum height, total velocity is not zero. Because velocity along x axis is present. Therefore velocity along x axis is, always remains constant throughout the motion. If you calculate velocity along x axis, Then every point is same.
But why so, This is because of, there is no any acceleration along x axis. And this indicate acceleration along x axis is zero. Hence velocity along x axis is constant. So using this all component velocity, and acceleration. You can find out projectile parameters by using trick.
- initial velocity along x axis = ucosα
- initial velocity along y axis = usinα
- acceleration along y axis = -g
- acceleration along x axis = 0
where u is velocity of projectile, and α is angle of projection with x axis. Now as i explain above . So by using straight line motion, in two separate axis x and y. You are able to find, all projectile parameter. Hence you can refer, straight line motion.
How do you explain 2D motion ?
So here are three equation. Which is given below
v = u+at, s = ut +1/2at² v² = u² +2as.
Now for projectile motion. You just use this equation to find its parameter .
So Time of flight is define, as total time taken by projectile in air. Hence You want to find time of flight.
Simply apply trick, you know that velocity along y axis is responsible, to kept in air. hence use velocity along y axis to find time of flight.
And you know that, initial velocity along y axis = usinα, and velocity along y axis at maximum height is zero. hence use this equation .
v = u +at where a = -g.
0 = usinα -gt after simplification t = usinα/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.
What is 2D motion explain its importance ?
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 v² = 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
How formula is calculate for 2D- projectile motion examples ?
R = ucos𝛳*time of flight – 1/2*0 = ucos𝛳*2usin𝛳/g = u²2sin𝛳cos𝛳/g = u²sin2𝛳/g hence R = u²sin2𝛳/g now for maximum range sin2𝛳 = 1 so 𝛳 = 45⁰
Q1 sometime asked question for which two value of 𝛳 Range will be same for this if one angle is 𝛳 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