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.
⇒ 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).