Show that the horizontal range is the same for the angles of projectio

1.	Show that the horizontal range is the same for the angles of projection 0 and (90^@-theta).
Answer» Solution :`R=(mu^2sin2theta)/g=(mu^2sin180^@-2theta)/g=(mu^2sin2(90^@-theta))/g`Hence for a given speed of projection, the HORIZONTAL RANGE is the .same for the ANGLES of projection 0 and `(90^@-theta)`

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