Projectile motion is the superposition of two independent motions: constant velocity horizontally and uniformly accelerated motion vertically. The key assumptions are: (1) acceleration due to gravity g is constant, (2) air resistance is negligible, (3) the range is small enough that Earth's curvature can be ignored.

The trajectory equation y = x*tan(theta)(1 - x/R) is particularly elegant for JEE. It shows the path passes through (0,0) and (R,0), with maximum height at x = R/2.

Complementary Angle Properties:

• R(theta) = R(90-theta): same range
• H$\frac{theta}{H}$(90-theta) = $tan^2$(theta): height ratio
• T$\frac{theta}{T}$(90-theta) = tan(theta): flight time ratio
• R = 4H*cot(theta): connects range and height

At the Highest Point:

• $v_y$ = 0 but acceleration = g (NOT zero)
• Speed is MINIMUM = u*cos(theta)
• KE fraction = $cos^2$(theta)
• Velocity is perpendicular to acceleration (angle = 90 degrees)
• Radius of curvature = $u^2$*cos^2$$\frac{theta}{g}

Projectile from Height h:

• Use quadratic: h + $u_y$t - $\frac{1}{2}$g$t^2$ = 0 (taking upward positive, ground at y = -h from launch)
• Two roots: positive one is physical