Two Dimensional Kinematics Question

[ayreia]

Homework Statement

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.

Homework Equations

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

 

[alphysicist]

Hi ayreia,

Homework Statement

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.

Homework Equations

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.

The quantity they are asking for is the magnitude of the total velocity at impact. But when you calculated 201m/s, you were not calculating the total velocity. What would you say the 201m/s is? Do you see how to get the answer they gave?

 

[ayreia]

Oh right, 201 m/s is only the y-component velocity, so I'll need to find the x-component as well, then use Pythagoras to find the total velocity?

 

[alphysicist]

Oh right, 201 m/s is only the y-component velocity, so I'll need to find the x-component as well, then use Pythagoras to find the total velocity?

That sounds good; and what is the x-component of the final velocity?

Using the Pythagorean theorem will get you the magnitude of the final velocity, and then you also need to find the direction. What do you get?