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Today I’m going to explain EVERYTHING about **Projectile motion formula**. As well as projectile motion examples.

Which you need to know, what is horizontal projectile motion, and projectile motion problem in this article.

So lets start from very basic concept.

**Projectile motion** best definition is here . Because projectile motion is easy, but conceptual. And this is best example of 2D motion. So very first, what is projectile motion and its equation equation ? . Hence it is define as, any body or particle having m mass but negligible size.

Therefore if you through any body vertically upward. Then that body moves in a straight line. Which is a vertical line. Similarly if you through a body, along horizontal line. Then again that body moves along horizontal straight line.

So both motion are straight line. But when you through a body between horizontal and vertical plan, making some angle with horizontal line. Then that body moves in a parabolic shape of path. So this is called projectile motion. And the path travel by projectile, is called **projectile motion equation.**

1What is the equation of projectile motion?

projectile motion is best example of 2D-motion, this motion occure in two plane (x,y). And acceleration only in the vertical direction. Which is called acceleration due to gravity (g). Equation of path of projectile motion is called trajectory. And path of this trajectory is y = (tan θ0)x – gx2/2(v0cosθ0)2 .With this equation, you’re able to calculate all parameter for projectile motion. for example maximum heigh, time of flight, velocity.

2What are examples of projectile motion?

Example If you throw a ball vertically upward, and it falls vertically downward, You kick a ball at an angle with horizontal with some speed. And let it fall, So all these’re example of projectile motion. Anything free fall is also under projectile motion.

3What are the three types of projectile motion?

1 Horizontal projectile, Which is any ball throw horizontal with some speed from certain height.2 Vertical projectile, This is anything free fall under gravity is called vertical projectile. 3 Oblique projectile, If anything throw at an angle with horizontal is type of oblique projectile.

This motion is also called free fall motion. Because when you project a body, then only gravitational force acts upon the body. And no any other forces act on the body. Hence this is called free fall motion.

Because all fundamental concept is applicable for **projectile motion**.

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.

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.

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.

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.

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.

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

→

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 sometime asked

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.

If you want to learn more topic of Physics. Then refer my previous post. This link is given below.

**What is Work best definition**?**Learn Physics best concept****What is escape velocity**?**Gravitational force and constant****Gravitational field intensity****What is Energy conservation**

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.

I'm Telecom Engineer by profession and Blogger by passion

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