Answer
$v_0^2 t^2= x^2 +(y+\dfrac{g \ t^2}{2})$
Work Step by Step
$x^2 +(y+\dfrac{g \ t^2}{2})=(v_0 \cos \alpha)^2 t^2+(v_0 \sin \alpha)^2 t^2$
or, $x^2 +(y+\dfrac{g \ t^2}{2})=v_0^2 t^2 ( \cos^2 \alpha+ \sin^2 \alpha)$
Since, $\cos^2 \alpha+ \sin^2 \alpha=1$
$x^2 +(y+\dfrac{g \ t^2}{2})=v_0^2 t^2 \times (1)$
or, $v_0^2 t^2= x^2 +(y+\dfrac{g \ t^2}{2})$
