Time it takes for an object to hit the ground

  • Thread starter AryRezvani
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  • #1
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Homework Statement



The height of a helicopter above the ground is given by h = 2.55t^3, where h is in meters and t is in seconds. At t = 1.65 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Homework Equations



The above equation, and d = 1/2at (because initial velocity is zero)


The Attempt at a Solution



2.55t^3
2.55(1.65)^3 = 11.45491875 m/s

11.45491875 = 1/2 (-9.8) (t^2)
22.9098375 = (-9.8) (t^2)
-2.3377 = t^2

You then get an imaginary number, but regardless, because time cannot be negative, I pretended like the answer was positive and took the square root.

I got uhhh 1.528966488, but when I plug it into the online homework, it's incorrect?
 

Answers and Replies

  • #2
631
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Homework Statement



The height of a helicopter above the ground is given by h = 2.55t^3, where h is in meters and t is in seconds. At t = 1.65 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Homework Equations



The above equation, and d = 1/2at (because initial velocity is zero)
You say the height of the helicopter (which contains the mailbag) is a function of time. The initial velocity therefore cannot be zero.
 
  • #3
rl.bhat
Homework Helper
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Since the h is the function of the time, the helicopter is rising. So when the mail bag is released, it's initial velocity is not equal to zero. Take the derivative of h and find the velocity of the mail bat at the time of release. It is in the upward direction.
 
  • #4
1,065
10
h(t) = 2.55t^3
It means the height changes with time.
You can find velocity v(t).
Initial velocity not zero.
 
  • #5
NascentOxygen
Staff Emeritus
Science Advisor
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The downdraft from the rotor would give the mailbag extra downward impetus. Are we to ignore this? :wink:
 
  • #6
utkarshakash
Gold Member
855
13
Hey. You are doing a very serious mistake. Since you have taken the downward direction as +ve(you took downward displacement of the ball to be positive in your equation), you must also take g to be +ve i.e. +9.8m/s. Also the ball's initial velocity wrt helicopter is 0 and not wrt ground. Take the time derivative of h and find the velocity of helicopter at 1.65 s. Then apply the concept of relative velocity to obtain the ball's velocity at the instant it is released. I have given you enough hints. Now you can solve this. ;)
 
  • #7
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Okay, initial velocity cannot be zero because the helicopter is constantly rising, correct?

So you take the first derivative of your position-time equation, and plug in the time given (1.65 sec) and you get 20.827125 m/s.

11.45491875 = (20.827125)(t) + 1/2(9.8)t^2

gravity being positive because we regarded the height (displacement) as positive, right?

Then you'd solve the above equation which i'm not sure how to do lol.
 
  • #8
rl.bhat
Homework Helper
4,433
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In the freely falling body, usually we take the ground as the reference point, or attach the co-ordinate axis to the starting point.
The displacement is the difference between the final position and the initial position.
In this problem it is negative. g is always in the downward direction. The quantities in the same direction must have the same sign. So rewrite your equation and solve the quadratic to find the time.
 
Last edited:

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