Two planes are each about to drop an empty tank. At the moment of the release each plane has the same speed of 135 m/s, and each tank is at the same height of 2.00 km above the ground. Although the speeds are the same, the velocities are different at the instant of release, because one plane is flying at an angle of 15.0 degrees above the horizontal (A) and the other is flying at an angle of 15.0 degrees below the horizontal (B). Find the magnitude and direction of the velocity with which the fuel tank hits the ground if it is from (a) plane A and (b) plane B. In each part, give the directional angles with respect to the horizontal.
Vf2 = Vo2 + 2ad
The Attempt at a Solution
Okay, so since I know how fast the plane is going, 135 m/s, I was able to calculate its speed for the x and y components, 130 m/s and 35 m/s respectively. I also have the displacement and acceleration for the fuel tank, so I figured I could just use the equation I referred to to solve it.
The square root of (35 m/s)2 + 2(-9.81 m/s)2(-2000 m) gave me approximately 201 m/s, but the answer I was given was 239 m/s, so I'm not really sure what I'm doing wrong here.
Also, I have no idea how to find the directional angles.
Thanks for the help :D