Physics (Free Falling Object)?

Question

The hight of a helicopter above the ground is given by h=3.00t^3 ,where h in meters and t in sec, After 2.00sec the helicopter releases a small mailbag. How long after it release does the mailbag reach the Ground?

Hint:. u can use following equations
x=v(i) t + 1/2 a t^2
v(f)^2=v(i)^2+2ax
V(f)=V(i)+at
a=9.8


If u no the answer , i realy want to thank u, coz i spent whole day i cant find the time?? , i checked the answer in the book, it says 7.9 sec, and i cant get that answer ><.
Thx all

Answer 1

I believe you have written down the problem incorrectly. I will show you the math but first stop and think about the physical
situation.
If h=3.00t^3 and t=2 sec, then the helicopter is only
24 meters off the ground. That is 76 feet.
Does it seem reasonable to you that something dropped from that height would take almost 8 seconds to reach earth?

No.
But if you like, we can put in the numbers.
x=1/2 at^2
24m=1/2 (9.8 m/sec^2)t^2
t^2=48/9.8 sec^2=4.9 sec^2
t=2.22 sec

So it took about 2.22sec for the package to fall.
Let's check.
In 2.22 sec the final velocity of the package would be
2.22sec(9.8 m/sec^2)=21.76m/sec
So the avg. velocity on the trip was 21.76/2=10.88m/sec
x=vt where tv is the avg. vel
so x=10.88(2.22)m=24.15m(the extra .15 comes from two previous roundings).
So with the information you have given, the helicopter was 24 m above the ground and the package took 2.22sec to fall.

One other possibility is that h=(3t)^3 , the whole quantity cubed.
If that is the case,h=6^3=216m
216=1/2(9.8)t^2
216/4.9sec^2=t^2
44.1=t^2
t=6.64sec

So, depending how the information in the problem is interpreted, t is either 2.22sec or 6.64sec.

So what about 7.9sec?
Let's work backwards and see how high the helicopter would have to be for the package to take 7.9 sec to reach the ground.( After 7.9 sec the avg vel of thepackage would be
7.9(9.8)/2=38.71m/sec
It had that vel for 7.9sec so the height is 7.9(38.71=305.8m
to reach that height the helicopter would have to travel
305.8/3=t^3sec^3
101.9=t^3
t=4.66sec (not the 2sec stated in the problem)

So you see, the package took either 2.22sec or 6.64sec to fall, not 7.9sec unless you wrote the problem down wrong.
Hope this helps. Remember to think of the physical situation you are dealing with to see if your answers make reasonable sense.

Answer 2

OK.

Step 1: Find the height when the mailbag is dropped.
t = 2.0 sec.
h(t) = 3*t^3
h(2) = 3*8 = 24m.

Step 2: Find the upward velocity of the helicopter when the mailbag is dropped:
h = 3t^3
v = dh/dt
dh/dt = 9t^2
v = 9t^2 = 9 * 4 = 36 m/s

Step 3: Given the initial height and the initial velocity, find the time it takes to hit the ground.
h = v(i)t + 1/2at^2
-24m = 36t + -4.9t^2 (it's -4.9 because the acceleration goes down, -24m because it's dropping 24 meters)
0 = -4.9 m/s^2 + 36t + 24

Quadratic equation:
t = (-b +/- (b^2 - 4ac)^1/2)/2a
t = (-36 +/- (1296 - -470.4)^1/2) / -9.8
t = 3.673 +/- (1766.4)^1/2 / -9.8
t = 3.673 +/- -4.289 sec, or -0.616 and 7.962 sec.

Ignore the negative outcome: t = 7.962 sec. (solution)

Answer 3

do the work till you get it!
this is cheating yourself! and you won't learn

Answer 4

please do your own work