Question

Chapter 09, Problem 013 A shell is shot with an initial velocity VO of 17 m/s, at an angle of 0o 51 with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (see the figure). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible? Exploion the

Answer 1

x = v_ix * t = [v_o² / g ] * sin 51 cos 51

= 17² * sin 51 cos 51 / 9.8 = 14.42 m

y = v_iy * t - [1/2 g t² ] = [v_o² / 2 g ] * sin² 51

= 17² * sin² 51 / (2 * 9.8) = 8.90 m

V_o = 2 v_o cos theta₀ = 2 * 17 * cos 51 = 21.39 m/s

t = sqrt [2y_i/g] = sqrt [2*8.9/9.8] = 1.35 s

x = x_i + v_i t = 14.42 + [21.38 * 1.35]

x = 43.2 m