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Class 10 Class 12
A projectile is projected with velocity u at an angle θ with horizontal. Discuss the trajectory of projectile. Also find the velocity of projectile at any instant.

Let us consider a projectile is projected with velocity 'u' at angle 'θ' with horizontal direction from a point on the ground.

Let the point of projection be origin, right direction is positive direction of X-axis and vertically upward direction is positive direction of Y-axis.

Equation of trajectory:

Break the components of velocity and acceleration due to gravity in horizontal and vertical direction.

x0 = 0    ; yo = 0
ux = u cos    ; uy = u sin
a= 0 ; ay = -g

Motion along horizontal direction:

...(1)

Motion along vertical direction:

...(2)

Eliminating t from (1) and (2),

This the required equation of trajectory.

Since the equation of trajectory is the equation of parabola, therefore projectile follows parabolic path.

Velocity of projectile:

Let at any instant, be the velocity of the projectile making an angle  with the horizontal.

i.e.,

ux = u cos  ,  uy = u sin

ax = 0 , ay = -g

Now,

vx = ux + axt    ,     vy =  uy + ayt

Therefore,

vx = u cos     , vy = u sin - gt

The velocity f the projectile is given by,

v =

Angle pf projection is given by, tan  =

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What is a vector quantity?

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