Projectile Motion Question

$v_t^2 - v_0^2 = 2as \\ v_t = \sqrt {2as + v_0^2} \\ = \sqrt {2as + (v \cdot cos {53^{\circ}})^2} \\ = \sqrt {2 \times 10 \times 40 + 6^2} \\ = \sqrt {836}$

$t = \frac {v_t - v_0}a = \frac {\sqrt {836} - 6}{10}$

$s = vt = v \cdot sin {53^{\circ}} \cdot t \\ = 10 \times sin {53^{\circ}} \times \frac {\sqrt {836} - 6}{10} \\ = 18.33$

Considering time and height relation

Or , $(-40-0) = - (10 cos 53°) t-5 t².$ (Take g=-10 m/s² , cos 53° =0.6 )

Or , $5t²+6t -40 =0$

      ≈t=2.291366 s

Distance covered = (velocity in horizontal direction )× $t$

      = 10 sin 53° × 2.291366 ≈  18.330932  (approx )