A ball is projected from the ground with a speed \( 15 \, \text{ms}^{-1} \) at an angle \( \theta \) with horizontal so that its range and maximum height are equal, then ‘\( \tan \theta \)’ will be equal to:
\( \dfrac{1}{4} \)
\( \dfrac{1}{2} \)
\( 2 \)
\( 4 \)