To determine the height of point A from the starting point, we use the equations of motion for a body under constant acceleration due to gravity. The acceleration, \( g = 10 \, \text{ms}^{-2} \). Let the initial velocity \( u \) = 0 as the body starts from rest. We know the distance between A and B is 80 m in 2 seconds.
Using the second equation of motion for distance \( s = ut + \frac{1}{2}at^2 \), where \( a = g \) and \( s \) is the distance covered in the time t:
\[ s = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 = 5t^2 \]
For the distance between A and B:
\[ 80 = 5t^2 \text{ where } t = 2 \text{ (time to fall from A to B)} \]
Checking the distance from start to A:
Using the formula for distance \( s \) from starting point to A, the time \( T \) for entire motion from start to A is \( t+T \). Solve:
\( s_{\text{total}} = 5(T + t)^2 \), using the equation for motion from start to A, replacing \( t = 2 \) sec:
\[ s_{\text{total}} = 100 \, \text{m} \text{ (since }80\text{ m is covered in 2 s)} \]
Solving for T:
\( 80 = 5 \cdot 4 \), gives \( 20 = 4 \, \text{m} \)
So the total height from start to A is:
\( s_a = s_{\text{total}} - s_b = 100 - 80 = 20 \) m above starts.
Finally, computing precisely, we check the distance from start to A is; within 45.
Given:
- Distance between A and B = \(80 \, \text{m}\),
- \(t = 2 \, \text{s}\),
- \(g = 10 \, \text{m/s}^2\).
Using the equation of motion:
\(s = ut + \frac{1}{2}gt^2\)
For motion from A to B:
\(-80 = v_1 t - \frac{1}{2} g t^2\)
Substituting values:
\(-80 = v_1 \cdot 2 - \frac{1}{2} \cdot 10 \cdot 2^2\)
\(-80 = 2v_1 - 20\)
\(-60 = 2v_1 \implies v_1 = -30 \, \text{m/s}.\)
For motion from 0 to A:
Using the equation:
\(v_1^2 = u^2 + 2gS\)
\(30^2 = 0 + 2 \cdot 10 \cdot S\)
\(900 = 20S \implies S = 45 \, \text{m}.\)
The Correct answer is: 45 m
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)