Question

obtain the expression for horizontal range of a projectile show that it is maximum at 45 degree​

Answer 1

Explanation:

nnnnxhxhxhxjxjx xbxjxjcju u BH uuxuxu hhhcbb

Answer 2

The average velocity of a projectile between the instant it crosses one third the maximum height. It is projected with u making an angle θ with the vertical.

There will be a pair of points for which vertical velocities at the same height are in opposite direction and therefore their average sum =0

It is the horizontal velocity which is uniform and hence v  

av  =u  x  =ucosθ

For a general point:

Displacement in Y-direction:

y=usinθ×t−  2 gt  2

Displacement in X-direction:

x=ucosθ×t

Now in order to calculate average velocity:

Average Velocity =NetDisplacement /TOTAL TIME

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