IIT (NTSE/Olympiad)  

Projectile Motion

Horizontal Projection

2. Horizontal projection
In Horizontal Direction In Vertical Direction
(i) Initial velocity ux = u
(ii) Acceleration = 0
(iii) Horizontal velocity of particle remains same after time t
horizontal velocity = vx= u
(iv) Range x = ut
(i) Initial velocity uy = 0
(ii) Acceleration = 'g' downward
(iii) Velocity of the particle after time t,
vy = 0 + (–g)t = –gt = gt (downward)
(iv) Displacement y = (1/2) gt2 (downward)
(v) Velocity at height h
vy2= 02 + 2(–h) (–g)
⇒ vy= √2gh
2.1 Velocity at a general point P(x, y) :
and

α is angle made by v with horizontal in clockwise direction
2.2 Displacement :
= x+ y

2.3 Trajectory Equation :

x = vxt = ut
y = – (1/2) gt2
eliminating t, we get
This is equation of trajectory
2.4 Time of flight:
– h = vyt – (1/2) gt2 = 0 – 1/2 gt2
t = ±
t = +
(negative time is not possible)
2.5 Range:
R = uxt = u
Note :
(i) If a projectile is projected with initial velocity u and another particle is dropped from same height at the same time, both the projectile would strike the ground with same vertical velocity. Both will have same vertical components of velocity but their net velocities would be different.

(ii) Relative motion of one projectile w.r.t. motion of particle dropped from same height at the same time would be in straight line joining them.
(iii) All the particles thrown with same initial velocity would strike the ground with same speed at different times irrespective of their initial direction of velocities.
(a) Time would be least for the particle thrown with velocity v downward i.e. particle (1)
(b) Time would be maximum for the particle thrown with velocity v vertically upwards i.e. particle (2)


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