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 bead P sliding on a frictionless semi-circular string... bead Q ejected... relation between $t_P$ and $t_Q$ is 
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