Elevator uniform speed problem

In summary: Let's let it die.In summary, the problem involves finding the height of an elevator at a given time T1 when a marble is dropped through the floor and falls with uniform acceleration g = 9.8 m/s^2, hitting the ground at time T2. Using the SUVAT equations, the height can be found in terms of T1 and T2, and with further manipulation, it can be expressed solely in terms of T1. The solution is given as h = (gT1(T2-T1)^2)/2T2. However, it is important to note that T2 is the time it took for the ball to hit the ground, not the final time of the event, as T2 is
  • #1
Mr Davis 97
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Homework Statement


At t = 0, an elevator departs from the ground with uniform speed. At time ##T_1## a child drops a marble through the floor.The marble falls with uniform acceleration g = 9.8 m/s^2, and hits the ground at time ##t = T_2##. Find the height of the elevator time ##T_1##.

Homework Equations


SUVAT Equations

The Attempt at a Solution


Using the equation ##h = v_e T_1## and ##0 = h + v_e T_2 - \frac{1}{2} g T_2^2##, I was able to find the height in terms of ##T_1## and ##T_2##: ##\displaystyle h = \frac{g T_2^2}{2(1 + \frac{T_2}{T_1})}##

However, this gives the height in terms of ##T_1## and ##T_2##, but I need it in terms of ##T_1##. How should I go about doing that?
 
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  • #2
You need to find time t for the object to reach the ground after being dropped.
t is just the difference between T1 and T2.
 
  • #3
I'm not seeing how I would find t...
 
  • #4
Mr Davis 97 said:

The Attempt at a Solution


Using the equation ##h = v_e T_1## and ##0 = h + v_e T_2 - \frac{1}{2} g T_2^2##, I was able to find the height in terms of ##T_1## and ##T_2##: ##\displaystyle h = \frac{g T_2^2}{2(1 + \frac{T_2}{T_1})}##

However, this gives the height in terms of ##T_1## and ##T_2##, but I need it in terms of ##T_1##. How should I go about doing that?
T1 and T2 are given: T1 is the instant of time when the child drops the marble and T2 is the time when it reaches the ground. The time of fall is T2-T1. You have to use the time of fall T2-T1 in the second equation.
 
  • #5
ehild said:
T1 and T2 are given: T1 is the instant of time when the child drops the marble and T2 is the time when it reaches the ground. The time of fall is T2-T1. You have to use the time of fall T2-T1 in the second equation.

So now I have ##\displaystyle h = \frac{gT_1(T_2- T_1)^2}{2T_2}##, but this doesn't seem to really get me anywhere...
 
  • #6
Mr Davis 97 said:
So now I have ##\displaystyle h = \frac{gT_1(T_2- T_1)^2}{2T_2}##, but this doesn't seem to really get me anywhere...
It is the solution. Both T1 and T2 are given data, and h is the elevator time at T1.
 
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Likes Mr Davis 97
  • #7
ehild said:
It is the solution. Both T1 and T2 are given data, and h is the elevator time at T1.
Ah! How did I miss that... Thanks!
 
  • #8
Mr Davis 97 said:
Ah! How did I miss that... Thanks!
:smile:
 
  • #9
s=ut+1/2at^2
h=0+1/2*9.8*(T2-T1)^2
initial velocity of the ball related to the elevator is zero.
 
  • #10
ehild said:
T1 and T2 are given: T1 is the instant of time when the child drops the marble and T2 is the time when it reaches the ground. The time of fall is T2-T1. You have to use the time of fall T2-T1 in the second equation.
T2 is the time it took the ball to hit the ground (T2 is being measured through a different clock and it was started the instant ball stated falling through the floor.) not the final time when the event end (starting point being when elevator started to rise).
 
  • #11
Mr Davis 97 said:
Find the height of the elevator time T1.
height of a time?
 
  • #12
Hkj_1029 said:
T2 is the time it took the ball to hit the ground (T2 is being measured through a different clock and it was started the instant ball stated falling through the floor.) not the final time when the event end (starting point being when elevator started to rise).
Not so. It is clearly stated that
Mr Davis 97 said:
At t = 0, an elevator departs from the ground … marble falls … and hits the ground at time ##t = T_2##.
Anyway, the thread is six years old.
 

Related to Elevator uniform speed problem

1. What is the "Elevator uniform speed problem"?

The "Elevator uniform speed problem" refers to a scientific problem in which an elevator moves at a constant speed without changing its acceleration or deceleration. This means that the elevator is moving at a uniform speed, which can be calculated using the distance and time traveled.

2. How is the uniform speed of an elevator calculated?

The uniform speed of an elevator can be calculated by dividing the distance traveled (in meters) by the time taken (in seconds). This will give the speed of the elevator in meters per second (m/s).

3. What are the factors that affect the uniform speed of an elevator?

The uniform speed of an elevator is affected by various factors such as the weight of the elevator, the friction between the elevator and the shaft, and the motor power of the elevator. Other external factors such as wind resistance and vibrations can also affect the uniform speed of an elevator.

4. How does the uniform speed of an elevator affect its energy consumption?

The uniform speed of an elevator can greatly affect its energy consumption. Moving at a constant speed requires a continuous supply of energy, so elevators that move at a uniform speed tend to consume more energy compared to elevators that have varying speeds.

5. What are the practical applications of the "Elevator uniform speed problem" in real life?

The "Elevator uniform speed problem" has practical applications in elevators, escalators, and other vertical transportation systems. Understanding and calculating the uniform speed of these systems is crucial for safety and efficiency in their operation. It can also be applied in other fields such as physics and engineering to better understand the principles of motion and energy.

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