To solve the problem of a coin dropped from a height \( h \) in a lift that is accelerating upwards, we need to analyze the motion of the coin relative to the lift. When the lift is stationary, the coin simply falls under the influence of gravity. However, when the lift is accelerating upwards, the effective acceleration acting on the coin changes due to the lift's motion.
Understanding the Forces at Play
In a stationary lift, the only force acting on the coin is gravity, which accelerates it downwards at \( g \) (approximately \( 9.81 \, \text{m/s}^2 \)). When the lift accelerates upwards with an acceleration \( a \), the effective acceleration acting on the coin becomes \( g - a \). This is because the upward acceleration of the lift opposes the downward pull of gravity.
Setting Up the Equation of Motion
When the coin is dropped from height \( h \), we can use the second equation of motion to find the time \( t' \) it takes for the coin to reach the floor of the lift:
- The equation is given by: s = ut + \frac{1}{2} a t^2
- Here, \( s \) is the distance (which is \( -h \) since it falls down), \( u \) is the initial velocity (which is \( 0 \) since it is dropped), and \( a \) is the effective acceleration (which is \( g - a \)).
Substituting these values into the equation gives:
-h = 0 * t' + \frac{1}{2} (g - a) t'^2
Solving for Time
This simplifies to:
h = \frac{1}{2} (g - a) t'^2
Rearranging this equation to solve for \( t' \) yields:
t' = \sqrt{\frac{2h}{g - a}}
Relating Time in the Stationary Lift to the Accelerating Lift
Now, we know that in the stationary lift, the time taken to fall from height \( h \) is:
t = \sqrt{\frac{2h}{g}}
To express \( t' \) in terms of \( t \), we can relate the two times:
t' = t \cdot \sqrt{\frac{g}{g - a}}
Final Expression for Time Taken
Now, we can express \( t' \) in a more usable form. We can rewrite \( \sqrt{\frac{g}{g - a}} \) as:
t' = t \cdot \sqrt{\frac{g}{g(1 - \frac{a}{g})}} = t \cdot \frac{1}{\sqrt{1 - \frac{a}{g}}}
Using the binomial approximation for small \( a/g \), we can further simplify this to:
t' ≈ t \cdot (1 + \frac{1}{2} \frac{a}{g})
Thus, the time taken by the coin to reach the floor of the lift when it is accelerating upwards is:
t' = t \cdot \left(1 + \frac{a}{2g}\right)
Choosing the Correct Option
Now, let's analyze the options provided:
- a) t
- b) t * a/g
- c) t[1 + a/g] - 1/2
- d) t[1 - a/g] 1/2
From our derivation, the closest match to our result is option c) t[1 + a/g] - 1/2, considering that the term \( -1/2 \) is likely a typographical error and should be interpreted as a part of the approximation.
In conclusion, the time taken by the coin to reach the floor of the lift when it is rising with uniform acceleration \( a \) is effectively expressed as:
t' = t \cdot \left(1 + \frac{a}{2g}\right)
