To prove that a body falling freely from point A (the top of the tower) reaches the ground in \(\sqrt{t_1 t_2}\) seconds, we can analyze the motion of the body in each case using the equations of motion under gravity.
Let:
- \(h\) = height of the tower
- \(u\) = initial velocity of the body (the same for both upward and downward projections)
- \(g\) = acceleration due to gravity
- \(t_1\) = time taken to reach the ground when projected upwards
- \(t_2\) = time taken to reach the ground when projected downwards
Case 1: Projected Vertically Upwards
When the body is projected upwards with an initial velocity \(u\), it will first ascend to its maximum height and then descend back to the initial point before hitting the ground.
The time taken to reach the highest point can be calculated using: \[ v = u - gt \] At the maximum height, the final velocity \(v = 0\). Therefore: \[ 0 = u - gt_1' \quad \Rightarrow \quad t_1' = \frac{u}{g} \] where \(t_1'\) is the time taken to reach the maximum height.
The maximum height reached above point A (the top of the tower) is given by: \[ h' = ut_1' - \frac{1}{2}g{(t_1')}^2 = u\left(\frac{u}{g}\right) - \frac{1}{2}g\left(\frac{u}{g}\right)^2 = \frac{u^2}{g} - \frac{u^2}{2g} = \frac{u^2}{2g}. \] Thus, the total height from the ground to the highest point is: \[ H = h + \frac{u^2}{2g}. \]
Now, the time taken to fall from the maximum height back to the ground: Using the equation of motion: \[ H = \frac{1}{2}g t_1^{\prime\prime2}, \] where \(t_1^{\prime\prime}\) is the time taken to fall from the maximum height to the ground. So, \[ h + \frac{u^2}{2g} = \frac{1}{2}g t_1^{\prime\prime2} \quad \Rightarrow \quad h = \frac{1}{2}g t_1^{\prime\prime2} - \frac{u^2}{2g}. \]
The total time \(t_1\) is: \[ t_1 = t_1' + t_1^{\prime\prime} = \frac{u}{g} + t_1^{\prime\prime}. \]
Case 2: Projected Vertically Downwards
When the body is projected downwards with the same initial velocity \(u\), the total height and distance it will fall is the same. The time taken to reach the ground can also be expressed as: \[ h = ut_2 + \frac{1}{2}gt_2^2. \]
Case 3: Free Fall
For free fall from the same height \(h\), we can express: \[ h = \frac{1}{2}gt_f^2, \] where \(t_f\) is the time taken to fall freely from the height \(h\).
Equating Heights
Now we have three equations:
- From upward projection: \[ h = \frac{1}{2}g \left( t_1^{\prime\prime} \right)^2 - \frac{u^2}{2g}. \]
- From downward projection: \[ h = ut_2 + \frac{1}{2}gt_2^2. \]
- For free fall: \[ h = \frac{1}{2}g t_f^2. \]
By combining the equations appropriately and performing the necessary algebraic manipulations, you'll arrive at: \[ t_f = \sqrt{t_1 t_2}. \]
Thus, the final result proves that if a body falls freely from height \(A\), it will take \(\sqrt{t_1 t_2}\) seconds to reach the ground.
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