Question

A projectile is projected at angle 60 with the horizontal.Elevation angle of the projectile at its highest point as seen from the point of projection is​

Answer 1

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Answer 2

Answer:

Elevation angle(α)= 1/ 2 tan

Explanation:

Let us consider a few things in order to tackle this question. Let assume the following for the shake of convenience;-

       u= velocity of projectile

      ∅= angle of projectile at the point of projection

       α= angle of elevation from point of projection to highest point.

      H= maximum height of the projectile

And, we know that, at the maximum height for a projectile, the horizontal range(R) becomes half i.e. R/2.

Now, refer to the attached image;

                tan α = H/ R/2

                tan α = 2H/ R

                tan α = 2 R tan ∅/ 4 R ( Since, H= R tan ∅/4 )

                tan α = 2/ 4  (since, tan 45= 1, given  ∅= 45 )  

                      α = 1/ 2 tan or tan inverse × 1/2

Hence, elevation angle of the projectile at its highest point as seen from the point of projection is​ α = 1/ 2 tan or tan inverse × 1/2.

Hope it helps ;-))