From a tower of height H, a particle is thrown vertically up with a speed V. The time taken by the particle to hit the ground 3 tiem that taken by it to reach the highest point of is path . Then find H.

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Define the Variables
Let:
- \( H \) = height of the tower
- \( V \) = initial velocity of the particle
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( t \) = time taken to reach the highest point
- \( t_f \) = total time of flight

### Step 2: Time to Reach the Highest Point
When the particle is thrown upwards, it will reach its highest point when its velocity becomes zero. We can use the first equation of motion:
\[ v = u + at \]
At the highest point, the final velocity \( v = 0 \). Therefore, we have:
\[ 0 = V - gt \]
Rearranging gives:
\[ gt = V \implies t = \frac{V}{g} \]
This is the time taken to reach the highest point.

### Step 3: Total Time of Flight
According to the problem, the total time of flight \( t_f \) is three times the time taken to reach the highest point:
\[ t_f = 3t = 3 \left(\frac{V}{g}\right) = \frac{3V}{g} \]

### Step 4: Displacement During the Total Time of Flight
Now, we will use the second equation of motion to find the total displacement:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, the displacement \( s \) is equal to the height of the tower \( H \) but in the downward direction, so we take it as negative:
\[ -H = V t_f - \frac{1}{2} g t_f^2 \]
Substituting \( t_f \):
\[ -H = V \left(\frac{3V}{g}\right) - \frac{1}{2} g \left(\frac{3V}{g}\right)^2 \]

### Step 5: Simplifying the Equation
Now, substituting \( t_f \) into the equation:
\[ -H = \frac{3V^2}{g} - \frac{1}{2} g \cdot \frac{9V^2}{g^2} \]
This simplifies to:
\[ -H = \frac{3V^2}{g} - \frac{9V^2}{2g} \]
Finding a common denominator:
\[ -H = \frac{6V^2}{2g} - \frac{9V^2}{2g} = \frac{-3V^2}{2g} \]
Thus, we have:
\[ H = \frac{3V^2}{2g} \]

### Final Result
The height of the tower \( H \) is given by:
\[ H = \frac{3V^2}{2g} \]

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