Projectile motion is a form of motion in which an object or particle (called a projectile) is thrown near the earth’s surface, and it moves along a curved path under the action of gravity only.**Motion along Horizontal direction: **A body projected from the ground with a velocity u at an angel with the horizontal.

Horizontal components of the velocity at time t is v_{x} = u cos θ

The horizontal distances covered in time t is x = (u cos θ) t

**Motion along Vertical direction: **Vertical components of the velocity at time t is v_{y} = u sinθ – gt

The vertical distance covered in time t is

y = (u sin θ) t – ½ gt^{2 }

**Equation of trajectory: **y = (tan θ) x – gx^{2} / 2u^{2} cos^{2}θ

It is in the form of y = Ax + Bx^{2}, which represents a parabola (where A = tan , B = – g / 2u^{2} cos^{2}θ)** **

**Resultant velocity at time t: **Magnitude of resultant velocity at time t is v = (v^{2}_{x} + v^{2}_{y}) ^{½}

The angle subtended by the resultant velocity vector with the horizontal is given by tan α = v_{y} / v_{x}

**Time of flight (T):** We know the general equation in vertical direction is y = (u sin θ) t – ½ gt^{2 }

For time of flight y=0

0= u sin θT – ½ gT^{2 }

U sinθ = gT/2

T = 2usinθ /g

**Maximum height attained: **We know the general equation in vertical direction is V^{2}_{y} – U^{2}_{y} = 2a_{y}s

For maximum height 0 – u^{2} sin^{2}θ = 2 (-g) h_{max}

h_{max} = u^{2} sin^{2}θ/2g

**Horizontal range: **We know the general equation in vertical direction is x = u cos θt

**For range t = T **

R = ucosθ * 2usinθ/g

R = u^{2} sin (2θ)/g.

**Applications**

**⇒ **The horizontal range is the same for angles θ and (90⁰ – θ).

**⇒ **The horizontal range is maximum for θ = 45⁰. R_{max} = u^{2}/g

**⇒ **When horizontal range is maximum, h_{max} = R_{max }/4

**⇒ **At the point of projection, KE = ½ mu^{2}, PE = 0. Total energy E = ½ mu^{2}.

**⇒ **At the highest point, KE = ½ mu^{2 }cos^{2}θ and PE = total energy – KE = ½ mu^{2} – ½ mu^{2 }cos^{2}θ = ½ mu^{2 }sin^{2}θ

**⇒ **To find R and from the equation of trajectory y = ax – bx^{2}

Where a and b are constants, refer to this figure.a) At O and B, y = 0. Putting y = 0 in the above equation, we have 0 = ax – bx^{2 }=> x = 0, x = a/b. Therefore R = a/b

b) At A y = h_{max} and x = R/2 = a/2b. Using these values in y = ax – bx^{2}, we get h_{max} = a^{2}/4b

**⇒ **If A and B are two points at the same horizontal level on a trajectory at a height h from the ground,a) t_{f }= 2usinθ/g = t_{1} + t_{2}

b) h = ½ g t_{1}t_{2}

c) Average velocity during time interval (t_{2} – t_{1}) is v_{av} = ucosθ (**·.·** during this interval, the vertical displacement is zero).