The direction (25 degrees north of east) is useless information. You can ignore it. If it is in the air 3.0 seconds,it takes 1.5 seconds to rise to maximum height. The vertical velocity component is g*1.5s = 14.7 m/s
To compute how far the arrow goes, you need to know the launch angle A from horizontal OR the horizontal velocity component. Neither has been provided.
b) 45 degrees. The range is
(V^2/g)*sin(2A) That is a maximum when A = 45 degrees.
An arrow directed at a stationary target is shot at an angle of [E 25 N]. The arrow hits the intended target at the same height from which it was launched, 3.0 s later.
a) Determine the distance from the archer to the target.
b) At what angle should the arrow be directed in to maximize its range?
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