The height of a helicopter above the ground is given by h = 3.40t^3 , where h is in meters and t is in seconds. After 2.05 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?
h = 3.40(2.05) = 29.3m
Using the derivative of height formula:
v(initial) = 10.2(2.05) = 42.9m/s
29.3 = 42.9t + 0.5(-9.8)t^2
solving the quadratic formula:
t=0.746s or t=8.01s
Is this correct?
Well, it seems we have a case of gravity vs. helicopter mailbags. Let's solve this in a lighthearted way, shall we?
To find out when the mailbag reaches the ground, we need to determine the height when it does. And when does that happen? When the height is zero!
So, let's set h = 0 and solve for t:
0 = 3.40t^3
Now, this equation is as stubborn as a mule, but we can tackle it! Divide both sides by 3.40 to get rid of that pesky coefficient:
0 = t^3
Hmm, the cube of time is equal to zero? That means the duration must be zero as well!
Therefore, the mailbag reaches the ground at the exact same moment it's released. It's like a magical teleportation trick for mailbags. Isn't that impressive?
Just remember, though, don't try this with your own mail. Let the professional helicopter mailbags handle it!
To find out how long after the mailbag reaches the ground, we need to determine when the height of the mailbag is equal to zero.
Given that the height of the helicopter above the ground is given by the equation h = 3.40t^3, we can substitute the value of h with zero to find the time when the mailbag reaches the ground.
0 = 3.40t^3
To solve for t, we divide both sides of the equation by 3.40:
0/3.40 = (3.40t^3)/3.40
0 = t^3
Since any number raised to the power of 3 is equal to 0 only when the number itself is 0, we can conclude that the mailbag reaches the ground when t = 0.
However, we need to consider when this occurs after 2.05 seconds, which is the time when the mailbag was released.
So, we add 2.05 seconds to the time when the mailbag reaches the ground, which is t = 0.
t = 0 + 2.05
t = 2.05 seconds
Therefore, the mailbag reaches the ground 2.05 seconds after its release.
How high is the chopper after 2.05 s?
call it H
H = 3.4 * 2.05^3 m
Now from that height, it drops something. I assume we're not taking air resistance into account here, or the question would be much longer. I don't know how accurate the measuure your class uses for g, the acceleration due to gravity. Usually it's 9.8 m/s, but sometimes a class uses 10.
So how long does it take to hit the ground from height H?
Interesting point: it depends on whether the chopper is still rising, which, according to the question, it is.
H = ut(m) + 0.5 a t(m) ^2 where u is initial velocity and a is acceleration, and t(m) is the time the mailbag is falling. In this case, a is g, and the initial velocity is opposite to the direction of acceleration, since the chopper is going up while the bag is going down.
From this, you can work out t(m), the time the mailbag is falling, if you can get u, the upward speed of the helicopter at 2.05 sec.
Is that enough?
oops...the displacement would have to be negative
therefore,
-29.3 = 42.9t + 0.5(-9.8)t^2
solving the quadratic formula:
t=-0.636s or t=9.39s
and the answer is t=9.39s
Sorry, missed your followup earlier.
I get the numbers you do, but something is funny with the question.
h = 3.40t^3
anything with a t^3 term isn't acceleration; it's jerk. Constant jerk, in fact, that would have the chopper well into outer space in 60 seconds! Sorry, I overlooked that earlier. Since it's jerk, its first derivative isn't velocity; you'd have to go to the second derivative to get velocity. Maybe the question didn't expect you to take the upward velocity of the helicopter into account. I don't know.