Given a Constant Acceleration magnitude of g/4, Find the value of t

In summary, the conversation revolves around solving for the time it takes for a sandbag to reach the ground after being released from a balloon. The quadratic equation 4gt^(2) - g(delta t)t - g(delta t) = 0 is used and the positive discriminate is taken since it deals with the passing of time. However, the resulting answer is wrong and the correct approach is being discussed. The velocity of the sandbag at release is also mentioned, and the importance of checking the signs in the final step is highlighted. Finally, the question of how much time it takes for the sandbag to reach the ground is addressed, with a note that the correct wording should be "how much time" instead of "how many seconds
  • #1
baumbad
3
0
Homework Statement
A hot-air balloon takes off from the ground traveling vertically with a constant upward acceleration of magnitude g/4. After a time interval Δt, a crew member releases a ballast sandbag from the basket attached to the balloon.
How many seconds does it take the sandbag to reach the ground? Express your answer in terms of Δt.
Relevant Equations
s = ut + 1/2at^(2)
I got to the quadratic equation of the motion where: 4gt^(2) - g(delta t)t - g(delta t) = 0 and tried to solve for t. In this case, we would take the positive discriminate since we are dealing with the passing of time.

t = ((sqrt(17) g(delta t)) + g (delta t)) / (8g)

However, this is the wrong answer and I am not sure why.
 
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  • #2
baumbad said:
4gt^(2) - g(delta t)t - g(delta t) = 0
Please show the steps by which you got that.
What was the velocity of the sandbag when it was released?
 
  • #3
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this is the work that I used, however, I'm not sure that this is the correct approach. At a particular time, the sandbag goes from an upward acceleration of g/4 to free fall.
 
  • #4
Your approach looks correct. You have two equations with arrows pointing at them. Check how you get from the first one to the second.
 
  • #5
Thank you very much for your help!
 
  • #6
Also check the signs in the final step.
 
  • #7
baumbad said:
After a time interval Δt, a crew member releases a ballast sandbag from the basket attached to the balloon.
How many seconds does it take the sandbag to reach the ground? Express your answer in terms of Δt.
Note that the question should have asked "how much time", not "how many seconds".

As written, it is a bit tricky to pick the right units for the required answer.
 
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Related to Given a Constant Acceleration magnitude of g/4, Find the value of t

What does "g" represent in the context of this problem?

"g" represents the acceleration due to gravity, which is approximately 9.8 meters per second squared (m/s²) on the surface of the Earth.

What is the initial condition required to solve for t?

To solve for t, we need to know the initial velocity (u) and the displacement (s) or the final velocity (v) of the object in motion under the given constant acceleration.

What formula is used to find the value of t under constant acceleration?

The formula used depends on the given conditions. Common formulas include:1. \( v = u + at \)2. \( s = ut + \frac{1}{2}at^2 \)3. \( v^2 = u^2 + 2as \)where \( a \) is the acceleration, \( t \) is the time, \( u \) is the initial velocity, \( v \) is the final velocity, and \( s \) is the displacement.

If the initial velocity is zero, how do you find t?

If the initial velocity \( u \) is zero and you know the displacement \( s \), you can use the formula \( s = \frac{1}{2}at^2 \). Given \( a = \frac{g}{4} \), you can solve for t as follows:\( t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{2s}{g/4}} = \sqrt{\frac{8s}{g}} \).

How does the value of g/4 affect the motion of the object?

The value of \( g/4 \) means the object is accelerating at one-fourth the rate of gravity. This lower acceleration results in a slower increase in velocity and a longer time to reach a given displacement compared to an acceleration of \( g \).

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