Falling Body Motion Principles

This question delves into the physics of objects in motion under the influence of gravity. Specifically, it requires calculating the total time a body takes to fall from a height, given partial information about its fall duration.

Falling Body Scenario Analysis

The problem provides the following details about the falling body:

• It starts from rest at the top of a tower.
• It covers half of the total height ($H/2$) in exactly $t_1 = 2$ seconds.
• The objective is to find the total time ($T$) required for the body to reach the ground from the top of the tower.

Physics Kinematics Equations

The motion described is one of constant acceleration due to gravity. The standard kinematic equation relating distance ($s$), initial velocity ($u$), acceleration ($a$), and time ($t$) is:

$$ s = ut + \frac{1}{2}at^2 $$

For an object falling from rest under gravity:

• Initial velocity, $u = 0$.
• Acceleration, $a = g$ (acceleration due to gravity).
• The equation simplifies to: $s = \frac{1}{2}gt^2$.

Calculating Fall Time Step-by-Step

Let $H$ represent the total height of the tower and $T$ be the total time taken to reach the ground.

1. Equation for Total Height: Using the simplified kinematic equation for the total distance $H$ covered in time $T$:

   $$ H = \frac{1}{2}gT^2 $$ (Equation 1)

2. Equation for Half Height: The body covers half the height ($H/2$) in $t_1 = 2$ seconds. Applying the same formula:

   $$ \frac{H}{2} = \frac{1}{2}g(t_1)^2 $$

   Substituting $t_1 = 2$ seconds:

   $$ \frac{H}{2} = \frac{1}{2}g(2)^2 $$

   $$ \frac{H}{2} = \frac{1}{2}g(4) $$

   $$ \frac{H}{2} = 2g $$ (Equation 2)

3. Solving for Total Time (T): We can substitute the expression for $H$ from Equation 1 into Equation 2. From Equation 1, we can express $H$ as $H = gT^2$. Substituting this into Equation 2:

   $$ \frac{gT^2}{2} = 2g $$

   To solve for $T$, first divide both sides by $g$ (assuming $g \ne 0$):

   $$ \frac{T^2}{2} = 2 $$

   Multiply both sides by 2:

   $$ T^2 = 4 $$

   There seems to be a calculation error in the previous step. Let's re-substitute correctly:

   From Equation 1: $H = \frac{1}{2}gT^2$. Substitute this $H$ into Equation 2: $\frac{H}{2} = 2g$.

   $$ \frac{1}{2} \left( \frac{1}{2}gT^2 \right) = 2g $$

   $$ \frac{1}{4}gT^2 = 2g $$

   Divide both sides by $g$:

   $$ \frac{1}{4}T^2 = 2 $$

   Multiply both sides by 4:

   $$ T^2 = 8 $$

   Taking the square root:

   $$ T = \sqrt{8} = 2\sqrt{2} \text{ seconds} $$

4. Numerical Approximation: Calculating the approximate value:

   Since $\sqrt{2} \approx 1.414$,

   $$ T \approx 2 \times 1.414 $$

   $$ T \approx 2.828 \text{ seconds} $$

Time of Fall Approximation

The calculated total time for the body to reach the ground is approximately $2.828$ seconds. This value closely matches one of the provided options.