Motion of any object in any of two axis involve (xy or yz or zx)

“Projectile Motion may be 2-D or even 3-D also but in general it is use to be 2-D Motion”



Vertical Mirror ®  Gravity acts in vertical direction so we can use equations of motion

Time taken by the image of ball going up = time taken going down = t (Let)

Use  v = u – g t

O = u sinq - g t

Total time



Now to get Maximum height

Using v2 = u2 – 2g h

(O)2 = (usinq)2 – 2g H


Velocity remains constant so we can not use equation of Motion

Note: Untill any external reason present to change ucosq, like air flow ucosq remains Constant

Now ucosq = Const


R = (u cosq) T


Now Sin2q = 2 sinq Cosq


Now we have







“These three results only when can be use if initial point of projection and final point of projection are at same level.

Example: A body of mass m is projected upward with initial velocity                                  then find time of flight, Maximum height attained and range attained by the body (g = 10m/s2)








Maximum range: To get maximum horizontal distance covered by any mass in projectile motion we must have a unique specified angle.


(Sin2q)max = 1

2q = 90°

q = 45°

To get maximum range angle of projection should be 45 °,

Note: Here we are neglecting the effect of air resistance.  If we Consider air resistance then this angle q should be little bit less than 45°


Mathematically there must be two different angle of projection for which we will get same range

Ball 1


Ball 2


Now if we assume

        a + b = 90° 

then b = 90° - a




If sum of angle of projection is = 90° and initial speed is same then both balls will have same range.

Note: Here in this Case only range will be same not time of flight and maximum height

R1 = R2

T1 ¹ T2

H1 ¹ H2

In case of same range relation of time of flights

Here we know a + b = 90°

Ball 1


Ball 2








In case of same range relation of maximum heights

Here again a +b = 90°

Ball 1


Ball 2










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