SOLUTION: A helicopter is taking off vertically when a crate slips out of the cargo door. At that instant, the helicopter is moving upward at 15 m/s and is 50 m above the ground. What is t

Click here to see ALL problems on Vectors

Question 83099: A helicopter is taking off vertically when a crate slips out of the cargo door. At that instant, the helicopter is moving upward at 15 m/s and is 50 m above the ground. What is the maximum height the crate will reach above the ground?
After falling from the helicopter, how long will the crate be in the air?
Please show work.
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
A helicopter is taking off vertically when a crate slips out of the cargo door. At that instant, the helicopter is moving upward at 15 m/s and is 50 m above the ground. What is the maximum height the crate will reach above the ground?
:
Using 1/2 the gravitational acceleration of a falling object; 9.8m/s^2:
h = -4.9x^2 + 15x + 50, where x = time in seconds
:
Find the axis of symmetry using x = -b/(2a); in this problem a=-4.9, b=+15
x = -15/(2*-4.9)
x = -15/-9.8
x = +1.53 seconds
:
Max height would be the vertex, find that, substitute 1.53 for x:
h = -4.9(1.53^2) + 15(1.53) + 50
h = -4.9(2.34) + 22.95 + 50
h = -11.47 + 22.95 + 50
h = 61.48 meters max height of the wayward box
:
:
After falling from the helicopter, how long will the crate be in the air?
That will be when h = 0
-4.8x^2 + 15x + 50 = 0
Use the quadratic formula; a=-4.9, b=15, c=50

:

:

:
Do the math here you should get a positive solution of
:
x = 5.1 seconds for it to hit the ground
:
A graph of this:

:
Did this help?