In a carnival, you have to throw a spear at an underwater target lying flat on t
ID: 2113514 • Letter: I
Question
In a carnival, you have to throw a spear at an underwater target lying flat on the bottom of a pool. The water is 1.0m deep. You're standing on a small stool that places your eyes 3.0m above the bottom of the pool. As you look at the target, your gaze is 30 degrees below horizontal. At what angle below horizontal should you throw the spear in order to hit the target? Your raised arm brings the spear point to the level of your eyes as you throw it, and over this short distance you can assume that the spear travels in a straight line.
Explanation / Answer
The situation is as shown in figure The angle of glance is i = 30 The depth of water is d = 1 m The height of the observer above the water is h = 3 m - 1 m = 2 m Refractive index of water is n = 1.33 Apply Snell's law at the water-air interface sin(90-i) = nsin(90-r) cosi = ncosr cos30 = (1.33) cosrr = 49.37 From the laws of triangle D = h/tani = 2m / tan30 = 3.464 m Also D' = d/tanr = 1m / tan49.37 = 0.85801231735m Therefore the angle at which the spear to be thrown is th = tan-1[(h+d)/(D+D')] th = tan-1[3.9 m/4.6 m] th = 34.7654077397 The situation is as shown in figure The angle of glance is i = 30 The depth of water is d = 1 m The height of the observer above the water is h = 3 m - 1 m = 2 m Refractive index of water is n = 1.33 Apply Snell's law at the water-air interface sin(90-i) = nsin(90-r) cosi = ncosr cos30 = (1.33) cosr
r = 49.37 From the laws of triangle D = h/tani = 2m / tan30 = 3.464 m Also D' = d/tanr = 1m / tan49.37 = 0.85801231735m Therefore the angle at which the spear to be thrown is th = tan-1[(h+d)/(D+D')] th = tan-1[3.9 m/4.6 m] th = 34.7654077397 From the laws of triangle D = h/tani = 2m / tan30 = 3.464 m Also D' = d/tanr = 1m / tan49.37 = 0.85801231735m Therefore the angle at which the spear to be thrown is th = tan-1[(h+d)/(D+D')] th = tan-1[3.9 m/4.6 m] th = 34.7654077397