# Projectile motion

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 vx = 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 vy = u sinθ – gt

The vertical distance covered in time t is

y = (u sin θ) t – ½ gt2

Equation of trajectory:    y = (tan θ) x – gx2 / 2u2 cos2θ

It is in the form of y = Ax + Bx2, which represents a parabola (where A = tan , B = – g / 2u2 cos2θ)

Resultant velocity at time t: Magnitude of resultant velocity at time t is v = (v2x + v2y) ½

The angle  subtended by the resultant velocity vector with the horizontal is given by tan α = vy / vx

Time of flight (T): We know the general equation in vertical direction is y = (u sin θ) t – ½ gt2

For time of flight y=0

0= u sin θT – ½ gT2

U sinθ = gT/2

T = 2usinθ /g

Maximum height attained: We know the general equation in vertical direction is V2y – U2y = 2ays

For maximum height 0 – u2 sin2θ = 2 (-g) hmax

hmax = u2 sin2θ/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 = u2 sin (2θ)/g.

Applications

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

The horizontal range is maximum for θ = 45⁰. Rmax = u2/g

When horizontal range is maximum, hmax = Rmax /4

At the point of projection, KE = ½ mu2, PE = 0. Total energy E = ½ mu2.

At the highest point, KE = ½ mu2 cos2θ and PE = total energy – KE = ½ mu2 – ½ mu2 cos2θ = ½ mu2 sin2θ

To find R and  from the equation of trajectory y = ax – bx2

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 – bx2 => x = 0, x = a/b. Therefore R = a/b

b) At A y = hmax and x = R/2 = a/2b. Using these values in y = ax – bx2, we get hmax = a2/4b

If A and B are two points at the same horizontal level on a trajectory at a height h from the ground,a) tf = 2usinθ/g = t1 + t2

b) h = ½ g t1t2

c) Average velocity during time interval (t2 – t1) is vav = ucosθ (·.· during this interval, the vertical displacement is zero).