Question

A projectleis fired from a c220 feet above the water at an inclination of 45 to the horizontal with a muzzle velocity of 45 feet per second. The heighth of the projectile above the water is given by h) +x220 where x is the horizontal distance of the projectile from the face of the cift Use this information to answer the following

Answer 1

d) h(x) = 32 m2 tnt 220 (452 A hoo (31.64, 235.82) (0,220 2-60 (-90.519) (153.80) The for the interval oenç 160 Connect graph is shown by He Dashed line Correct figure in option | (c) (e) when h(a)= 100 feet To find x - 32x² 100 +x+220 こ - 32n² ta + 120 = 0 452 -1 + J (02- yx ca). na - x = * () x = 124.34 22-61.06 Neglected rounding off x= 124 feet

Solution: given h(x) = - 32x² (45) tx+220 (b) for maximum height of the projectile dhn) du au l 32n² x +220) (45) - 64 a + 1 = 0 x - (45) -64 = 31.64 x= 31.64 2 Henie Max height h (31-64) +31.64 +220 -32 x 81:64) (452 -15.82 +31.64+ 220 235.82 15.82 +220 = h (31064) = max height Max height = 235.82) feet- 236 feet (C) When the pojecht will stile wicter surface then hm-o - -4x220 + or -32n² (2452 tnt 220 = 0 (را) x = 153.80 we get Solving x= 154 feel -90.51 Neglected Any