v ( initial ) = 20 m/s h = 2.30 m h = v y * t + g t ² / 2 d = v x * t 1 ) At α = 18°: v y = 20 * sin 18° = 6.18 m/s v x = 20 * cos 18° = 19.02 m/ s 2.30 = 6.18 t + 4.9 t² 4.9 t² + 6.18 t - 2.30 = 0 After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ): t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 ) t = 0.3 s d 1 = 19.02 m/s * 0.3 s = 5.706 m 2 ) At α = 8°: v y = 20* sin 8° = 2.78 m/s v x = 20* cos 8° = 19.81 m/s 2.3 = 2.78 t + 4.9 t² 4.9 t² + 2.78 t - 2.3 = 0 t = 0.46 s d 2 = 19.81 * 0.46 = 9.113 m The distance is: d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m
