Projectile motion: Velocity and trajectory
Understand the velocity components and trajectory equation in projectile motion with clear derivations and visual insights. Perfect for NEET, JEE Main, and JEE Advanced aspirants aiming for conceptual clarity and accuracy in physics.
Velocity:
A ball is thrown with a velocity 
at an angle 
with the horizontal. Neglecting air resistance, find the time after which its velocity becomes perpendicular to the velocity of projection .
Let the velocity of the ball become perpendicular to the velocity of projection after time t and let this velocity be 
. Then, = + 
Since these two velocities are mutually perpendicular, we get . = 0
( + ). = 0
Or, v2 + vgtcos(900 + 
) = 0
Or, v2 – vgtsin = 0
t = 
Thus, after a time of , the velocity is perpendicular to the velocity of projection.
Some concepts on equation of trajectory of projectile motion:
1. Trajectory of projectile motion using range of a projectile: When a particle is projected at an angle with horizontal with speed u, then the equation of trajectory of the particle is y = xtan – 
g(
)2
If R is the range of the particle then R = 
.
Therefore, y = xtan – 
Or, y = xtan – 
Or, y = xtan – 
Or, y = xtan – 
y = xtan[1 – 
].
2. A particle is projected in x-y plane from the origin O with an angle with positive x axis. The particle moves in a parabolic path OAB in x-y plane. The line OA and AB create angles 
and 
respectively on x axis as shown in figure. Derive a relation between the angles , and .
Let us consider, the co-ordinate of point A is (a, h) and that of point B is [(a+b), 0].
Then, tan = 
and tan = 
From equation of trajectory we get, y = xtan[1 – ].
Or, h = atan[1 – 
] [range R = a+b]
Or, = tan[
]
Or, tan = (
) = + = tan + tan
∴ tan = tan + tan.
3. The equation of the trajectory of a projectile is y = ax – bx2 where a and b are constant and x and y are the horizontal and vertical displacement of the particle at any instant. Calculate the maximum height of the trajectory and the angle of projection with horizontal.
When a particle is projected with speed u at an angle with horizontal then the trajectory of a projectile is y = xtan – 
g(
)2.
Comparing the above equation with y = ax – bx2 we get, a = tan and b = 
or, u2 = 
.
Maximum height H = 
= 
= 
= 
.
Angle of projection = tan-1(a).
4. Two particles are projected from a point in x-y plane with speed u1 and u2 and angle 
and 
respectively with positive x axis. What is the nature of the projectile of 2nd particle observed with respect to 1st particle?
Let us consider, at time t, two particles are at point P (x1, y1) and Q (x2, y2) respectively.
Then x1 = u1cos and y1 = u1sint – 
Similarly, x2 = u2cos and y2 = u2sint – .
The position of 2nd particle with respect to the 1st particle are
x = x2 – x1 = u2cost – u1cost
And y = y2 – y1 = u2sint – u1sint
Therefore, 
= 
= m (constant)
Therefore, the nature of the projectile of 2nd particle observed with respect to 1st particle is a straight line (y = mx).
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