A projectile is fired horizontally. Show that the path followed by the object is parabolic Obtain the expression for the horizontal range and time of flight.

a. y = (tanθ) x - 1/2gx² / (v₀cosθ) ²

b. Range R = v₀²sin2θ/g

c. Time of flight T = 2v₀sinθ/g

Explanation:

a. Show that the path followed by the object is parabolic

Let the vertical component of its velocity be v₁ = v₀sinθ and the horizontal component of its velocity be v₂ = v₀cosθ where v₀ = initial velocity of the projectile.

Now, for its vertical displacement Δy = v₁t - 1/2gt² (1)

Its horizontal displacement Δx = v₂t (2)

From (2) t = Δx/v₂

Substituting t into (1), we have

Δy = v₁ (Δx/v₂) - 1/2g (Δx/v₂) ²

= (v₁/v₂) Δx - 1/2g (Δx) ² / (v₂) ² =

= (v₀sinθ/v₀cosθ) Δx - 1/2g (Δx) ² / (v₀cosθ) ²

If we assume the initial position is at the origin, then Δx = x - 0 = x and Δy = y - 0 = y. So,

y = (tanθ) x - 1/2gx² / (v₀cosθ) ².

This is the required equation and it is a quadratic equation in x. Thus, the path followed by the projectile is parabolic.

b. The Range, R

The range, R is obtained when y = 0

So, 0 = (tanθ) x - 1/2gx² / (v₀cosθ) ²

[ (tanθ) - 1/2gx / (v₀cosθ) ²]x = 0

x = 0 or (tanθ) - 1/2gx / (v₀cosθ) ² = 0

x = 0 or (tanθ) = 1/2gx / (v₀cosθ) ²

x = 0 or (tanθ) (v₀cosθ) ² = 1/2gx

x = 0 or x = 2 (sinθ/cosθ) (v₀cosθ) ²/g

x = 0 or x = v₀² (2sinθ/cosθ) / g

x = 0 or x = v₀²sin2θ/g [sin2θ = 2sinθ/cosθ]

So its range R = x = v₀²sin2θ/g

c. Time of flight, T

The time of flight, T is obtained from (1) when Δy = 0 and t = T

So, 0 = v₁T - 1/2gT²

(v₁ - 1/2gT) T = 0

T = 0 or (v₁ - 1/2gT) = 0

T = 0 or v₁ = 1/2gT

T = 0 or T = 2v₁/g = 2v₀sinθ/g

So, the time of flight T = 2v₀sinθ/g