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

photocells convert light energy to electricity

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000