An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers?

The plane is traveling horizontally at 100 m/s at a height of 50.0 m above the ground.

1. What horizontal distance does the package travel before striking the ground?

2. Find the velocity (magnitude and direction) of the package in the previous question just before it hits the ground.An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers?Assume negligible air resistance.

First calculate the time it takes for the package to drop using the equation: h = gt^2/2

where h = 50 m, g = 9.8 m/sec^2

So t = sq rt (2h/g)

1) The horizontal distance is the horizontal component of the velocity which is 100 m/sec multiplied by the time that you found.

2) The vertical component of the velocity is v = gt

The magnitude of the the final velocity is the square root of the sum of the squares of the two components of the velocity.

To get the direction, draw a vector diagram. The angle the velocity makes with the ground is the arctangent of the vertical component divided by the horizontal component.An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers?To find the horizontal distance, you need to know how long the package stays in the air.

This teaching video on a falling object should help, but it takes to part 3 (part 4 is OK too) to get the variables you have and need:

http://www.khanacademy.org/video/project鈥?/a>

http://www.khanacademy.org/video/project鈥?/a>

http://www.khanacademy.org/video/project鈥?/a>

http://www.khanacademy.org/video/project鈥?/a>An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers?This is a projectile problem so i suggest breaking things down into x an y components. To find the displacement in x, you need velocity and time. You have velocity but you don't have time so use the y components to find time first. Realize that time is the only thing that's always similar in x and y.

x y

Vi=100m/s Vi=0 m/s d=Vit + 1/2at^2.If you use this equation to solve for time,you get 3.19 s.

d=-50 m

a=-9.8 m/s^2

Once you have found time, use the equation v=d/t to find you displacement in x. You should get 31.3 m.

Now part 2 is asking for the final velocity in y. Use your givens in y to figure out which kinematics equations would work for your problem. In this case, either Vf=Vi + at, or Vf^2=Vi^2 + 2ad would work. The first one is probably the easier one though. Anyways, using either of those equations, your final velocity should come out somewhere around 31.3 m/s. The direction would be down.