Question

Proof of tan theta =tan alpha + tan beta in projectile motion

Answer 1

Answer:

In order to proof tanθ = tanα + tanβ in a projectile motion we are required to take certain assumptions.  

Explanation:

Consider a 2-dimensional x-y plane with AB as the horizontal base along the x-axis. A particle is thrown from a point of origin A (0,0) as shown in the figure, at an angle “θ”, known as angle of projection, and with a velocity of projection “u”, which touches the ground or base at point B of AB forming an angle “β” . The particle forms a triangle, ∆ABC, with vertex C (x, y).

Let the horizontal range, that the particle covers along the base be “R”

Time of flight be “T”

Time taken by the particle to reach C (x, y) be “t”= distance/speed= x/ucosθ

The velocity of projection along the x-axis = u cosθ

The velocity of projection along the y-axis = u sinθ

Now draw a perpendicular CD or “y” from vertex C (x, y) to the base AB at point D as shown in the figure attached below. Therefore, let AD = x and  DB = (R-x).

Thereby considering the particle be at motion under gravity we can deduce  

y = u t + ½ at²  

or, y = u sinθ t – ½ gt² …………………(i)

[ since the force of gravitation is always working downwards for vertical motion, therefore acceleration along the vertical axis is “-g”.]

x = u t + ½ at²  

or, x = u cosθ t + ½ (0)t²  

or, x= u cosθ t……………. (ii)

[ since the no horizontal force is present therefore the horizontal acceleration is zero.]

Solving for equation (i) & (ii)

y= {u sinθ *  (x/u cosθ)} – {(1/2 * g * x²)/(u²cos²θ)}

or, y= [u sinθ * { x/(u cosθ)}] – {( g * x² * sec²θ)/(2 u²)}

or, y/x = tanθ – {(g sec²θ) / (2 u²)}x…………………. (iii)

Since the vertical displacement is zero during the time of flight "T" therefore,  

0 = u sinθ T – ½ g T²

or, 0 = T{u sinθ - (T g/2)}

or, 0 = {2u sinθ - g T} / 2

or, T = (2 u sinθ)/g

Now the horizontal displacement at the time of flight which is the range “R” will be  

x= u cosθ T

or, x= u cosθ * {(2 u sinθ)/g}

or, x= (u² sin2θ) / g

or, R={2 u²tanθ} / {g(1+tan²θ)}.........[ since sin2θ = tanθ/(1+tan²θ)]

or, R={2 u²tanθ}/{g sec²θ}

or, {g sec²θ} / {2 u²} = tanθR …………………… (iv)

Solving the equation (iii) and (iv)  

y/x=tanθ − {(1/2*g * x * sec²θ) / (u²)}

or, y/x = tanθ − {(x tanθ) / R}

or, tanα = tanθ − {(x tanθ) / R} ......  [since yx = tanα from the figure in ∆ADC ]

So tanθ = tanα * {R / (R−x)}

Or, tanθ = tanα* {(R−x+x) / (R−x)}

Or, tanθ = tanα * [ 1 + {x / (R−x)}]

Or, tanθ = tanα + {(x tanα) / (R−x)}

Put x tanα = y

Or, tanθ = tanα + {y / (R−x)}

From the figure below, considering ∆BDC we can get

{y / (R−x)} = tanβ

Hence,  

tanθ = tanα + tanβ