1-D,2-D and 3-D motion, Trajectory(path) of projectile
1-D MOTION
2-D MOTION OR MOTION IN A PLANE Þ
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”
-D PROJECTILE MOTION
VERTICAL MIRROR
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
HORIZONTAL MIRROR
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
CONDITION
“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)
Solution
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°
SAME RANGE
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
Now
In case of same range relation of maximum heights
Here again a +b = 90°
Ball 1
Ball 2
=
Path of Projectile
TRAJECTORY (PATH) OF PROJECTILE
Along y- axis using
Along x-axis using
By this equation it can be prove that path of projectile is parabola
OTHER FORMA OF EQUATION OF TRAJECTORY
Projectile Motion when initial point and final point are at different level
Case I Horizontal projection from certain height
Here angle of projection q = 0˚
In vertical Mirror
Using
Time to strike ground in horizontal Mirror
IN HORIZONTAL MIRROR
=
Example: Which ball will strike the ground first
(a) Ball A (c) Ball C
(b) Ball B (d) All Balls will strike simultaneously
Solution
There fore all three Balls strike the ground simultaneously Answer (d)
Example: A jet fighter plane is flying at constant speed of 360 km/h and jet plane is moving horizontally at a height of 20 km from earth ‘s surface. Jet release a bomb to hit a target on the ground, how much horizontal gap before jet will release the bomb
Solution
As discussed above using
Example: A stair case having each step 10 cm high and 20 cm wide.
If a ball drops in horizontal direction from the top step with 10 mt/sec
in horizontal direction If ball strikes nth step of staire case find the n?
Solution
Now make a twist in above question
Every data is same except velocity of ball, if initial velocity not given , and saying what will be minimum initial velocity in above question so that ball hits 50th step
Solution
Now x = 49 x 0.2 mt
h = 49 x 0.1 mt
Projection in upwards and downwards directions
Case II Projection in upward direction with some angle with horizonta
Vertical mirror
sign convention
Horizontal Mirror
x = (u cosq) t
Case III Projection in downward direction with some angle with horizontal
Vertical Mirror
Horizontal Mirror
Projectile along inclined plane For Simplicity & understanding
Vertical Mirror
V = u - g t
O = u sin b - (g cosa) t
Horizontal Mirror
OB = OA Cos a
Special case of Projectile Motion
Case I
(i) Time of flight let OQ ® x -axis
OP ® y –axis
ux = u
uy = 0
ax = - g sin b
ay = - g cos b
Now
Vx = ux + axt
O = u – g sin bt
(ii) Final velocity V
Vy = uy + ay t
V = 0 – g cos b
V = u cot b
(iii) Height h
Now
h = Sy sin a
(iv) Distance AB
Now
Case 2
(i) Time after which V1 ^ V2
Ball 1
Ball 2
Now
g2 t2 = u1 u2
(ii) gap between the ball 1 and ball 2 at the time
Ball 1 x1 = u1 t,
Ball 2
x2 = - u2t
Gap between Ball 1 and Ball 2
Gap = (u1 + u2) t
Case 3 Collision of two projectile