Physics Problems: Fighter Jet, Forces, and Airplane Speeds Explained

[goodie2shoes]

Hello. I do not really understand how to solve these problems and was hoping that you could help/explain them to me. Thanks.

1. A fighter plane moving 200 m/s horizontally fires a projectile with speed 50.0 m/s in a forward direction 30 degrees below the horizontal. What is the speed of the projectile with respect to a stationary observer on the ground?
I was thinking to use cosθ = A/H. But that doesn't work. How do I set up this problem?!

2. Three forces, each having a magnitude of 30N, pull on an object in directions that are 120 degrees apart from each other. What is the resultant?

3. An airplane with a speed of 120km/h is headed 30 degrees east of north in a wind blowing due east at 30km/h. What is the speed of the plane relative to the ground?
I tried using cos, sin/ tan, but somehow I can't get the answer again.

4. Two forces are acting upon an object as shown. What is the magnitude of the resultant forces.?

So basically I kind redid the diagram here. a is 75 degrees between the left hand side lines (slashes) and the box and b is 60 degrees for the right hand side lines (slashes) and the box.

80N...\.../...120N
...\.../
...a .\./...b
... ----------
...|...|
...
I was going to use F and such to solve for the resultant but no mass is given and I can't think of another way to solve for it.

Thanks for your help. Much appreciated.

 

[member 553083]

1. To solve this problem, you can set up a right triangle with the horizontal and vertical components of the projectile's velocity. The horizontal component will be 50 m/s * cos(30°) = 43.3 m/s, and the vertical component will be 50 m/s * sin(30°) = 25.0 m/s. You then add the two components together to get the total speed of the projectile relative to the ground: 43.3 m/s + 25.0 m/s = 68.3 m/s. 2. To solve this problem, you can use the law of cosines. Let A = 30N, B = 30N, and C = 30N. Then, the magnitude of the resultant force can be found using the equation: C^2 = A^2 + B^2 - 2*A*B*cos(120°). Plugging in the given values, we get: 30^2 = 30^2 + 30^2 - 2*30*30*cos(120°), or 30^2 = 900 - 900*(-0.5), or 30^2 = 900 + 450 = 1350. Taking the square root of both sides, we get that the magnitude of the resultant force is √1350 = 36.7 N. 3. To solve this problem, you can use trigonometry. The plane's speed relative to the ground can be found by adding the plane's speed (120 km/h) and the wind's speed (30 km/h) together, but in different directions. The plane's speed can be broken down into a horizontal and vertical component, and the wind's speed can also be broken down into a horizontal and vertical component. The horizontal components will be in the same direction, and the vertical components will be in the same direction, so you can just add them together. The horizontal component of the plane's speed is 120 km/h * cos(30°) = 104 km/h, and the vertical component is 120 km/h * sin(30°) = 60 km/h. The horizontal component of the wind's speed is 30 km/h, and the vertical component is 0. So, the total horizontal component of the plane's speed is 104 km/h + 30 km/h = 134 km/h, and the total vertical component is 60 km/h