Magnitue and direction of velocity

[pink1]

1. 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 tak 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, and the other is flying at an angle of 15.0 degrees below the horizontal. Find the magnitude and direction of the velocity with which the fuel tank hits the ground if it is from plane A(15 degrees above the horizontal), and plane B(15 degrees below horizontal). In each part, give the directional angles with respect to the horizontal.

2. Homework Equations : X= Xo+Vot+1/2 at^2 (same for y)
Vx= Vo+at (same for y)


I found my time to be 408 seconds (2 Xdistance in m(2000m)/a(9.8 m/s)
I then added my x and y coodinates for plane A:X=130, Y=35; for plane B: X=-130, y=-35. For the most pary I am lost. I'm not sure of how to find my y velocity or if I need to find it. Please help!

 

[mgb_phys]

First you need to resolve the planes speed into a horizontal and vertical component.
Assuming no air resistance the horizontal componnet of the tank is constant so you need to use the vertical component to find the time and hence the vertical components of final velocity.

 

[pink1]

thanks for the help,
so my speed would be 13.8s. I took my initial speed 135- my initial velocity 0, and divided by gravity. but that does not seem right

 

[mgb_phys]

No if your speed is 135mph at 15degrees your horizontal speed is 135 cos(15) = 130 horizontal and 135 sin(15) 35mph up/down.
Draw a triangle with the long side 135units and an angle of 15 degrees to see why.

 

[huda_1]

hi i got the velocity being 239m/s but not being able to find the directions of the plane??