A projectile is fired from a submarine traveling horizontally at 20 m/s with respect to the water as shown in the figure below. According to an observer on the submarine, the projectile is fired at 45° with an initial velocity of 60 m/s. After firing the projectile, the submarine continues to travel at 20 m/s.

According to an observer watching from a boat that is stationary with respect to the water, what will be the angle θ that the projectile makes with respect to the horizontal when it is launched?

(a) θ = 30.9°
(b) θ = 34.2°
(c) θ = 45°
(d) θ = 62.1°
(e) θ = 71.6°

http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/oldexams/exam1/sp10/fig14.gif

Did you mean to put this in physics, not psychics? :)

Yes, I select the wrong option.

1. Break the velocity of the projectile according to an observer on the submarine (ie, don't take into consideration the speed of the submarine for now)

2. The apparent vertical velocity is the same as seen by an observer on the boat.

3. The apparent horizontal component is however something else. It is the horizontal velocity component of the projectile, minus the speed of the submarine (since they are going in opposite direction).

4. Take those two apparent velocities, and use the tangent ratio to find the apparent angle.

And since the projectile appears to move less along the horizontal, you'll have an angle greater than 45 degrees.

Can you solve this?