__Question 4.32__:

(a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

θ(t) = tan

^{-1}((v

_{0y}- gt)

**/**v

_{0x})

(b) Shows that the projection angle θ

^{0}for a projectile launched from the origin is given by

θ

_{0}= tan

^{-1}(4h

_{m}

**/**R)

where the symbols have their usual meaning.

__Solution__:

**(a)**Let v

_{0x}and v

_{0y}respectively be the initial components of the velocity of the projectile along horizontal (

*x*) and vertical (

*y*) directions.

Let v

_{x}and v

_{y}respectively be the horizontal and vertical components of velocity at a point P.

Time taken by the projectile to reach point P =

*t*

Applying the first equation of motion along the vertical and horizontal directions, we get:

v

_{y}= v

_{0y}= gt

And v

_{x}= v

_{0x}

∴ tan θ = v

_{y}/ v

_{x}= (v

_{0y}- gt)

**/**v

_{0x}

θ = tan

^{-1}(v

_{0y}- gt) / v

_{0x}

**(b)**Maximum vertical height, h

_{m}= u

_{0}

^{2}Sin

^{2}θ

**/**2g ...(i)

Horizontal range, R = u

_{0}

^{2}Sin

^{2}2θ

**/**g ... (ii)

h

_{m}/ R = Sin

^{2}θ

**/**2Sin

^{2}2θ

= Sin θ X Sin θ

**/**2 X 2SinθCosθ

= Sin θ

**/**4 Cos θ = tan θ

**/**4

tan θ = (4h

_{m}/ R)

θ = tan

^{-1}(4h

_{m}/ R)

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