You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. Unfortunately the archer stands on an elevated platform of unknown height. However, you find the arrow stuck in the ground 67.0m away, making a 2.00 degree angle with the ground.
How fast was the arrow shot?
Hint: In this question, you do not need to know the initial y-position of the arrow, nor should you assume a value. . The final direction of the arrow indicates the ratio of the v_y/v_x. So tan(final_angle with respect to the x-axis)= v_y/v_x but neither v_y, nor v_x are known. Since the arrow initially was fired horizontally, it only has an x component of velocity at first, and since there is no acceleration in x, v_x final= v_x initial. So you only need to find v_x final to get the initial velocity. You need another equation that relates v_y and v_x. Well, v_y = -gt and you can get t from v_x*t=(displacement in x). Use the last two equation to eliminate t, which is not required for this problem, which give another equation that involves v_y and v_x.
? m/s
Dear Student, Please find the solution attached herewith. Regards
Download Preview:
Horizontal distance traversed by arrow, s = 67 m
Horizontal component of velocity be vx m/s
Vertical component of velocity be vy m/s
tan2o = -vy/vx
also, vy = -g * t
where t is time for which arrow...