A body is thrown vertically up reaches a maximum height of 78.4 m. After what time it will reach the ground from the maximum height reached ?

Options

A. 4 sec
B. 8 sec
C. 3 sec
D. 5 sec

Answer

To solve the problem of finding the time taken for a body to reach the ground from its maximum height of 78.4 m, we can use the kinematic equations of motion. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the problem
The body is thrown vertically upwards and reaches a maximum height of 78.4 m. At this maximum height, the final velocity (v) is 0 m/s. We need to find the time (t) it takes for the body to fall back to the ground from this height.

### Step 2: Identify the known values
- Maximum height (h) = 78.4 m
- Initial velocity at maximum height (u) = 0 m/s (since the body is momentarily at rest at the peak)
- Acceleration due to gravity (g) = 9.8 m/s² (acting downwards)

### Step 3: Use the kinematic equation
We can use the kinematic equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s \) = displacement (which is the height fallen, 78.4 m)
- \( u \) = initial velocity (0 m/s at maximum height)
- \( a \) = acceleration (which is -g, but we will use g as positive since we are considering the downward motion)
- \( t \) = time taken to fall

Substituting the known values into the equation:
\[ 78.4 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]

### Step 4: Simplify the equation
Since the initial velocity \( u \) is 0, the equation simplifies to:
\[ 78.4 = \frac{1}{2} \cdot 9.8 \cdot t^2 \]

### Step 5: Solve for \( t^2 \)
Multiply both sides by 2 to eliminate the fraction:
\[ 2 \cdot 78.4 = 9.8 \cdot t^2 \]
\[ 156.8 = 9.8 \cdot t^2 \]

Now, divide both sides by 9.8:
\[ t^2 = \frac{156.8}{9.8} \]

### Step 6: Calculate \( t \)
Now, calculate \( t^2 \):
\[ t^2 = 16 \]
Taking the square root of both sides gives:
\[ t = \sqrt{16} = 4 \text{ seconds} \]

### Final Answer
The time taken for the body to reach the ground from the maximum height is **4 seconds**.

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