question_answer

A ball is projected with velocity u at an angle \[\alpha \] with horizontal plane. What is its speed when it makes an angle \[\beta \] with the horizontal plane?

Answer:

                Let \[\upsilon \] be the velocity of ball at an instant, when it makes an angle P with the horizontal. The horizontal component velocity of the ball = \[\upsilon \cos \beta \]. Initial horizontal component velocity of ball = \[u\cos \alpha \] In angular projection of a projectile, the horizontal component velocity remains unchanged, hence \[\upsilon \cos \beta =u\cos \alpha \]    or                            \[\upsilon =u\cos \alpha /\cos \beta \]