Question
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.
Solution
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
ax = 0 ; ay = -g
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


ax = 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 =