Question

1. Bull's-Eye Bonanza. (Allotted Time 45 minutes) Suppose you want to throw a dart across a room at a wall-mounted target and hit the target exactly at its center point. We call this center point of the target its "bull's eye". The bull's-eye is a horizontal distance d from the dart's point of release and a height h above the dart's point of release. To hit the bul'seye, you have a choice of the speed s at which you throw the dart anda chotce of the angle e above the horizontal direction at which you release the dart. In other words, e is the angle between the dart's initial velocity vector and the horizontal direction. For any angle at which you choose to throw the dart, it can be shown that there is at most one speed that will allow it to hit the bull's-eye. If you throw the dart too softly, it will land under the bul'seye, and if you throw the dart too hard, it will land over the bull's-eye

Answer 1

The equation in the first row second column is the answer the reason is

The path that should be followed by the dart is Parabolic Projectile. The equation of Parabolic Projectile is y = x tan theta - [1/2 g/v^2 cos^2 theta] x^2 here y = h, x = d theta = theta, v = v, theta = theta, g = g substituting. h = d tan theta - [g/2 1/v^2 cos^2 theta]d^2 h - d tan theta = -g/2 d^2/cos^2 theta 1/v^2 \r\n = > 1/v^2 times gd^/2 cos^2 theta = d tan theta - h = > 1/v^2 = (d tan theta - h) 2 cos^2 theta/gd^2 = > v^2 = gd^2/2 cos^2 theta(d tan theta - h) = > v = d/cos theta squareroot g/2(d tan theta - h) = > v = d/cos theta squareroot g/2/d tan theta - h