) A striker can give the ball an initial speed of 30m/s. Within what two elevation angles must he kick the ball to score a goal from a point 50m in front of the goalposts whose horizontal bar is 2.44m
above the ground?
(

  about 19.6° and 73.2°

Explanation:

The equation for ballistic motion in Cartesian coordinates for some launch angle α can be written ...

  y = -4.9(x/s·sec(α))² +x·tan(α)

where s is the launch speed in meters per second.

We want y=2.44 for x=50, so this resolves to a quadratic equation in tan(α):

  -13.6111·tan(α)² +50·tan(α) -16.0511 = 0

This has solutions ...

  tan(α) = 0.355408 or 3.31806

The corresponding angles are ...

  α = 19.5656° or 73.2282°

The elevation angle must lie between 19.6° and 73.2° for the ball to score a goal.

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I find it convenient to use a graphing calculator to find solutions for problems of this sort. In the attachment, we have used x as the angle in degrees, and written the function so that x-intercepts are the solutions.