A boy is standing on a hill inclined at angle @ to the horizontal. He throws the ball at angle of elevation 45 degrees and speed v m/s. If he can throw the ball 60m down the hill but only 30m up the hill, use the cartesian equation of the path y = x - (gx^2)/(v^2) to show that:
tan @ = 1 - (30gcos@)/(v^2) = (60gcos@)/(v^2) - 1
Any solutions would be greatly appreciated.
tan @ = 1 - (30gcos@)/(v^2) = (60gcos@)/(v^2) - 1
Any solutions would be greatly appreciated.