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Question
Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by
`theta(t) =tan^(-1) ((v_(0y) - "gt")/v_(o x))`
Solution 1
Let `v_(ox)` and `v_(oy)` 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_(oy) = "gt"`
And `v_x =v_(ox)`
`:. tan theta = v_y/v_x= (v_(oy) - "gt")/v_(ox)`
`theta = tan^(-1)((v_(oy) - "gt")/(v_(o x)))`
Solution 2
Let the projectile be fired at an angle `theta` with x-axis.
As `theta` depend on t, `theta(t)` at any instance
`tan theta(t) = v_y/v_x =(v_(oy) - "gt")/(v_(o x))`
(Since `v_y = v_(oy) - "gt" and v_x = v_(o x)`)
`=> theta(t) = tan^(-1)((v_o - "gt")/v_(o x))`
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