The direction of a projectile at a certain instant is inclined at an angle `prop` to the horizontal , after `t` second, it is inclined at an angle `beta`. Prove that the horizontal component of the velocity of the projectile is `("gt")/(tan prop - tan beta)`.

Explanation

Let the projectile be at point `P` at any instant, with its velocity inclined at an angle `prop` to the horizontal.
Therefore, horizontal component `= u cos prop` …(i)
and vertical component `= u sin prop`...(ii)
where `u` is the velocity of the projectile at `P`.
Let after `t` seconds, the particle reaches point `Q`, with angle of inclination `beta` to the horizontal and velocity `v`.
Resolving velocity at `Q`,
Horizontal component `= v cos beta` ...(iii)
and vertical component `= v sin beta` ...(iv)
Therefore, horizontal component remains same.
`:. u sin prop = v cos beta` ...(v)
And for the vertical motion from point `P to Q`,
`u = u sin prop , v sin beta , a = -g , t = t`
Using `v = u + at`
`v sin beta = u sin prop - "gt"`....(vi)
Substituting for `v` from Eq. (vi) in Eq. (v),
`u cos prop = ((u sin prop - "gt")/(sin beta)) cos beta = ( u sin prop - "gt") cos beta`
`u sin prop . cot beta - u cos prop = "gt". cot beta`
`u cos prop [tan prop. cot beta - 1] = "gt". cot beta`
`u cos prop = ("gt" cos beta)/(tan prop. cot beta - 1) = ("gt")/(tan prop - tan beta)`
[Multiplying Nr. and Dr. by `tan beta`]
Therefore, horizontal component,
`u cos prop = ("gt")/(tan prop - tan beta)`.
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