A body freely falling from a height h describes `(7h)/16` in the last second of its fall. The height h is `(g = 10 ms^(-2))`

Options

A. 80 m
B. 45 m
C. 160 m
D. 40 m

Answer

To solve the problem, we need to find the height \( h \) from which a body is freely falling, given that it covers \( \frac{7h}{16} \) in the last second of its fall. We will use the equations of motion under gravity.

### Step 1: Understanding the problem
We know that the body falls freely under the influence of gravity, which means its motion can be described using the equations of motion. The distance fallen in \( n \) seconds can be expressed as:
\[ h = \frac{1}{2} g t^2 \]
where \( g = 10 \, \text{m/s}^2 \).

### Step 2: Finding the time of fall
Let \( t_0 \) be the total time of fall. The distance fallen in the last second (i.e., from \( t_0 - 1 \) to \( t_0 \)) can be calculated using the formula:
\[ \text{Distance in last second} = \frac{1}{2} g t_0^2 - \frac{1}{2} g (t_0 - 1)^2 \]

### Step 3: Simplifying the distance formula
Substituting the values into the distance formula:
\[ \frac{1}{2} g t_0^2 - \frac{1}{2} g (t_0^2 - 2t_0 + 1) = \frac{1}{2} g (2t_0 - 1) \]
So, the distance fallen in the last second can be expressed as:
\[ \text{Distance in last second} = \frac{1}{2} g (2t_0 - 1) \]

### Step 4: Setting up the equation
We know that this distance is equal to \( \frac{7h}{16} \). Therefore, we can write:
\[ \frac{1}{2} g (2t_0 - 1) = \frac{7h}{16} \]

### Step 5: Expressing \( h \) in terms of \( t_0 \)
From the equation \( h = \frac{1}{2} g t_0^2 \), we can substitute \( h \):
\[ \frac{1}{2} g (2t_0 - 1) = \frac{7}{16} \left(\frac{1}{2} g t_0^2\right) \]
Cancelling \( \frac{1}{2} g \) from both sides (assuming \( g \neq 0 \)):
\[ 2t_0 - 1 = \frac{7}{16} t_0^2 \]

### Step 6: Rearranging the equation
Rearranging gives us:
\[ \frac{7}{16} t_0^2 - 2t_0 + 1 = 0 \]

### Step 7: Multiplying through by 16 to eliminate the fraction
Multiplying through by 16:
\[ 7t_0^2 - 32t_0 + 16 = 0 \]

### Step 8: Solving the quadratic equation
Using the quadratic formula \( t_0 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t_0 = \frac{32 \pm \sqrt{(-32)^2 - 4 \cdot 7 \cdot 16}}{2 \cdot 7} \]
Calculating the discriminant:
\[ t_0 = \frac{32 \pm \sqrt{1024 - 448}}{14} = \frac{32 \pm \sqrt{576}}{14} = \frac{32 \pm 24}{14} \]

### Step 9: Finding the possible values for \( t_0 \)
Calculating the two possible values:
1. \( t_0 = \frac{56}{14} = 4 \) seconds
2. \( t_0 = \frac{8}{14} = \frac{4}{7} \) seconds (not physically meaningful as time cannot be negative)

### Step 10: Finding the height \( h \)
Now substituting \( t_0 = 4 \) seconds back to find \( h \):
\[ h = \frac{1}{2} g t_0^2 = \frac{1}{2} \cdot 10 \cdot (4^2) = \frac{1}{2} \cdot 10 \cdot 16 = 80 \, \text{meters} \]

### Final Answer
The height \( h \) is \( 80 \, \text{meters} \).

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