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Motion In A Plane

Question
CBSEENPH11019965

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

                              straight theta left parenthesis straight t right parenthesis thin space equals space tan to the power of negative 1 end exponent space open square brackets fraction numerator straight v subscript oy space minus space gt over denominator straight v subscript ox end fraction close square brackets

Solution
Let v0x and v0y respectively be the initial components of the velocity of the projectile motion along horizontal (x) and vertical (y) directions. 

Let vx and vy be the horizontal and vertical components of velocity at a point P respectively. 
Time taken by the projectile to reach point P = t
Using the first equation of motion along the vertical and horizontal directions, we get
vy = v0y = gt, and
vx = v0x
therefore space tan space straight theta space equals space straight v subscript straight y over straight v subscript straight x space equals space fraction numerator open parentheses straight v subscript 0 subscript straight y space end subscript minus space gt close parentheses over denominator straight v subscript 0 straight x end subscript end fraction

rightwards double arrow space straight theta space equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator straight v subscript 0 subscript straight y space end subscript minus space gt over denominator straight v subscript 0 straight x end subscript end fraction close parentheses