you are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. you ask one of the archers to shoot an arrow parallel to the ground. you find the arrow stuck in the ground 72m away, making a 5.0 deg angle with the ground. assume that the arrow always points in thesame direction as its velocity. How fast was the arrow shot?

Choose the coordinate system such that the arrow is shot from the origin, the x-axis is directed to the right, and the y-axis is directed downwards. Let t be the flight time of the arrow and (x, y) be the coordinates of its landing point. The initial velocity is (v_x; 0) and the final velocity is (v_x; v_y).

Given x = 72 m; v_y / v_x = tan θ = tan 5°. Find v_x.

Use the equations: x = v_x * t; y = g * t^{2}/2 where g = 9.81 m/s^{2}, and v_y = g*t.

Since tan θ = v_y / v_x = g*t / (x/t) = g * t^{2 }/ x, we find that t = sqrt{x * tan θ / g) and

v_x = x / t = sqrt{ g * x / tan θ) = sqrt{ 9.81 * 72 / tan 5°} = 90 m/s (with 2 significant figures).