Question

Ian 410109cle055d6pittmath1%2fdefault 127308 O E Messages Courses Help Logout dent section: 18978) NALYTC GEOMETRY & CALCULUS 3 Timer Notes Evaluate Feedback Print a Info Course Contents PROBLEM SET 03 projectile motion PROJECTILE A projectile is fired with an initial muzzle speed 510 m/s at an angle 45 from a position 6 meters above the ground leve Find the horizontal displacement from the firing position to the point of impact. Use g 9.8m/s. meters Submit Answer Incorrect. Tries 3/8 Previous Tries At what speed does the projectile hit the ground? meters/second 806.47 Submit Answer Incorrect. Tries 2/8 Previous Tries Send Feedback Post Discussion

Answer 1

initial horizontal component of velocity Vx = V cos 45 = 510 * cos 45 = 360.57 m/s

initial vertical component of velocity Vy = V cos 45 = 510 * sin 45 = 360.57 m/s

use for vertical motion Y = Vyt -0.5 gt^2

4.9t^2 -360 .57 t -6 = 0

solving for t , t = 73.60 secs

now for horizontal motion, distance S = Vxt = 360.57 * 73.06

Sx = 26.53 km -----------<<<<<<<<<<<<Answer to part A

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vxf = u -gt = (360.57 - 721.23) = - 360.73 m/s

Vf^2 = Vx^2 + Vy^2

Vf^2 = 360.73^2 + (-360.73^2)

Vf = 510 m/s --------------------------<<<<<<<<<<<Answer to part B