Solving for the Alaskan Rescue Team Equation

[Let It Be]

1. An Alaskan rescue team drops a package of emergency rations to a stranded hiker. The plane is traveling horizontally at 40 m/s at a height of 100m above the ground. A) Where does the package strik the ground relative to the point at which it was released? B) What are the horizontal and vertical components of the velocity of the package just before it hits the ground?


2. H=1/2gt^2
ΔX=ViTf+1/2ATf^2
AND apparently ΔY=ViTf+1/2ATf^2


3. So I'm not quite sure why you can't use the free fall equation...that was my first guess but it just seems too easy. The book says to use the third equation I put above, however I don't know how to completely solve it.
I got this far...
ΔY=ViTf+1/2ATf^2
100=?(would the Vi be 40m/s??)Tf+1/2(9.8)Tf^2
I know I'll end up with a Tf & Tf^2 on the right side of the equation, but I don't know what to do to solve.

Right, and then part B I'm soooo lost

 

[cupid.callin]

1. An Alaskan rescue team drops a package of emergency rations to a stranded hiker. The plane is traveling horizontally at 40 m/s at a height of 100m above the ground. A) Where does the package strik the ground relative to the point at which it was released? B) What are the horizontal and vertical components of the velocity of the package just before it hits the ground?2. H=1/2gt^2
ΔX=ViTf+1/2ATf^2
AND apparently ΔY=ViTf+1/2ATf^23. So I'm not quite sure why you can't use the free fall equation...that was my first guess but it just seems too easy. The book says to use the third equation I put above, however I don't know how to completely solve it.
I got this far...
ΔY=ViTf+1/2ATf^2
100=?(would the Vi be 40m/s??)Tf+1/2(9.8)Tf^2
I know I'll end up with a Tf & Tf^2 on the right side of the equation, but I don't know what to do to solve.

Right, and then part B I'm soooo lost

100m is vertical or horizontal?
40m/s is vertical or horizontal velocity?
PS: This ques is just like rolling marble ques.

 

[Let It Be]

100m is vertical or horizontal?
40m/s is vertical or horizontal velocity?
PS: This ques is just like rolling marble ques.

100m is vertical
40m/s is horizontal
Could you help me figure out how to solve the ΔY=ViTf+1/2ATf^2 part please

P.P.S-Really? In that one I had to find Vi too...

 

[cupid.callin]

100m is vertical
40m/s is horizontal
Could you help me figure out how to solve the ΔY=ViTf+1/2ATf^2 part please

P.P.S-Really? In that one I had to find Vi too...

here also you have it

initial horizontal velocity is same as of _____
and initial vertical is 0 as it is just dropped

 

[asteriatic]

.

I would approach this problem by breaking it down into smaller, more manageable parts. First, let's define some variables:

- t: time in seconds
- v0: initial velocity (in this case, the horizontal velocity of the plane, which is 40 m/s)
- g: acceleration due to gravity (9.8 m/s^2)
- h: height of the plane above the ground (100 m)
- x: horizontal distance traveled by the package
- y: vertical distance traveled by the package

Now, let's tackle part A of the problem. We want to find where the package will strike the ground relative to the point at which it was released. This means we need to find the value of x when y = 0. We can use the equations you provided to help us solve for x:

- ΔX=ViTf+1/2ATf^2
- AND apparently ΔY=ViTf+1/2ATf^2

Let's start with the second equation, since it already has y in it. We can substitute in the values we know:

0 = v0t + 1/2gt^2

Now, we can rearrange this equation to solve for t:

t = -v0/g

We can then plug this value for t into the first equation:

x = v0(-v0/g) + 1/2g(-v0/g)^2

Simplifying, we get:

x = -1/2g(-v0^2/g)

Now, we know that -v0^2/g is just the height of the plane (h), so we can substitute that in:

x = -1/2gh

And since we want the value of x when y = 0, we know that x = 0. So, the package will strike the ground at a horizontal distance of 0 from the point at which it was released, meaning it will hit directly below the plane.

For part B, we need to find the horizontal and vertical components of the package's velocity just before it hits the ground. This can be done using the same equations, but this time we want to find the values of v0x (horizontal velocity) and v0y (vertical velocity). We can use trigonometry to find these values:

v0x = v0cosθ
v0y = v0sin