How Does Package Dropping Dynamics Work for Airborne Deliveries?

[05holtel]

Homework Statement

A relief airplane is delivering a food package to a group of people stranded on a very small island. The island is too small for the plane to land on, and the only way to deliver the package is by dropping it. The airplane flies horizontally with constant speed of 290 mph at an altitude of 850 m. For all parts, assume that the "island" refers to the point at a distance D from the point at which the package is released, as shown in the figure. Ignore the height of this point above sea level. Assume that the acceleration due to gravity is g = 9.80 m/s^2.

a) After a package is ejected from the plane, how long will it take for it to reach sea level from the time it is ejected? Assume that the package, like the plane, has an initial velocity of 290 mph in the horizontal direction.

b)If the package is to land right on the island, at what horizontal distance D from the plane to the island should the package be released?

c)What is the speed v_f of the package when it hits the ground in mph
Equations

x1=x0 + v0x(t1-t0)
y1= y0 + v0y(t1-t0) -0.5(g)(t1-t0)^2

The Attempt at a Solution

a)13.1s
b)1710m
c) I am not sure how to do c. At first i thought speed would be 0 when it hits the ground, but i am told that is incorrect

 

[tiny-tim]

The airplane flies horizontally with constant speed of 290 mph at an altitude of 850 m. For all parts, assume that the "island" refers to the point at a distance D from the point at which the package is released, as shown in the figure. Ignore the height of this point above sea level. Assume that the acceleration due to gravity is g = 9.80 m/s^2.
…
c)What is the speed v_f of the package when it hits the ground in mph
Equations
…
I am not sure how to do c. At first i thought speed would be 0 when it hits the ground, but i am told that is incorrect

Hi 05holtel!

Hint: v_f² = v_horizontal² + v_vertical²

 

[05holtel]

vf² = vhorizontal² + vvertical²

Does that mean
Vf^2 = 290mph^2 +0mph^2
Vf = Square root of (290^2)
Vf=290mph

 

[tiny-tim]

vf² = vhorizontal² + vvertical²

Does that mean
Vf^2 = 290mph^2 +0mph^2
Vf = Square root of (290^2)
Vf=290mph

erm … v_vertical = 0 when it leaves the plane.

use the usual formula to find v_vertical 13.1s later.

 

[05holtel]

So does that mean:
Vfy=Viy-g(Delta t)
Vfy=0-4.38(13.1) ***9.8m/s = 4.38mph
Vfy=751.65mph

Vfx=0mph

Therefore, Vf = 751.65
Is that right?

 

[tiny-tim]

So does that mean:
Vfy=Viy-g(Delta t)
Vfy=0-4.38(13.1) ***9.8m/s = 4.38mph
Vfy=751.65mph

Vfx=0mph

Therefore, Vf = 751.65
Is that right?

No, Vfx= 290mph.

Vfy=Viy-g(∆t) is right.

But your g is wrong.

g is not 9.8m/s … it's 9.8m/s² … you need to convert it to miles/s².

 

[05holtel]

So does that mean:

9.8m/s^2 = 0.006 089 mile/s^2

Vfy=290mph-0.006089mile/s^2(13.1s)
Vfy=290mph-0.7976m/s
Vfy=290mph-1.178mph
=288.8mph

 

[tiny-tim]

So does that mean:

9.8m/s^2 = 0.006 089 mile/s^2

Yes.

Vfy=290mph-0.006089mile/s^2(13.1s)
Vfy=290mph-0.7976m/s
Vfy=290mph-1.178mph
=288.8mph

i] 290 is Vfx.

ii] it's 0.07976m/s

iii] where did 1.178 come from?

 

[05holtel]

OOPs


Vfy=290mph-0.006089mile/s^2(13.1s)
Vfy=290mph-0.7976mile/s
Vfy=290mph-2871.36mph
=?

That doesn't make any sense. Am I converting it wrong