Question

Example: Shooting an Arrow A good compound bow can shoot an arrow with an initial velocity of 100 m/s. If you shot this bow at an angle of 30° above the horizon, what would be the arrow's position and velocity after 2.5 seconds?

Answer 1

a) v^vector = v_ox i^^+ (v_oy + a_y t)j^^= 100 cos 30 degrees i^^+ (100 sin 30 degrees - 9.8 times 2.5)j^^v^vector = 86.6 m/s i^^+ 25.5 m/s j^^velocity after 2.5s V = squareroot (86.6)^2 + (25.5)^2 = 90.3 m/s direction theta = tan^-1[25.5/86.6] = 16.4 degrees counter clockwise from + x-axis b) r^vector = xi^^+ yj^^from x = v_ox t + 1/2 a_x t^2 x = (100 cos 30 degrees) times 2.5 + 0 x = 216.5 m from y = v_oy t + 1/2 a_y t^2 = 100 times sin 30 degrees times 2.5 - 1/2 times 9.8 times 2.5^2 = 94.375 m position r^vector = (216.5 m)i^^+ (94.4 m)j^^Magnitude r = squareroot (216.5)^2 + (94.4)^2 r = 236.2 m