Question 57837This question is from textbook College Algebra A Graphing approach
: A cargo plan is flying at an altitude of 30,000 feet and a speed of 540 miles per hour (792 per second. How many feet will a supply crate dropped from the plane travel horizontally before it hits the ground if the path of the crate is modeled by
x = -39,204(y – 30,000)?
This question is from textbook College Algebra A Graphing approach
Answer by phillv(13)   (Show Source):
You can put this solution on YOUR website! I am not sure what this equation means. However this physics kinematics problem can be solved using vectors.
The distance an object travels when an acceleration is involved can be calculated using the formula:
S=1/2at^2 where S is distance, a is acceleration and t is time of travel.
For this problem we are dropping an object from a plane.
There is a vertical component (vector) and a horizontal component (vector) involved.
The plan is to:
1. Find the time the crate will take to hit the ground (vertical)
2. Use the time to calculate how far the crate will travel in the given time (horizontal)
We can substitute acceleration due to gravity (g) for (a) since the crate is falling and rearange the equation to solve for (t).
t=sqrt(2S/g) where S is vertical distance the crate drops.
t=sqrt(2(30000 ft)/32 ft/(s^2))
t=sqrt(60000 ft)/32 ft/(s^2))
t=sqrt(1875 s^2)
t= 43.3 s
It will take 43.3 seconds for the crate to hit the ground.
Since we know the horizontal speed of the crate when it left the plane:
792 ft/s
We can use the formula S=vt where S is distance, v is velocity and t is time of travel.
S=(792 ft/s)43.3 s
S=342936 ft
The crate will land a horizontal distance of 34936 ft from where it was droped.
|
|
|
