A body thrown with an initial speed of `96 ft//sec` reaches the ground after `(g = 32 ft//sec^(2))`

Options

A. 3sec
B. 6sec
C. 12sec
D. 8 sec

Answer

To solve the problem of a body thrown upwards with an initial speed of 96 ft/sec and reaching the ground under the influence of gravity (g = 32 ft/sec²), we can use the kinematic equation for motion:

### Step-by-Step Solution:

1. **Identify the Variables**:
- Initial velocity (u) = 96 ft/sec (upward)
- Acceleration (a) = -g = -32 ft/sec² (downward, hence negative)
- Displacement (s) = 0 ft (the body returns to the ground level)

2. **Use the Kinematic Equation**:
The kinematic equation we will use is:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the known values:
\[ 0 = 96t + \frac{1}{2}(-32)t^2 \]

3. **Simplify the Equation**:
This simplifies to:
\[ 0 = 96t - 16t^2 \]
Rearranging gives:
\[ 16t^2 - 96t = 0 \]

4. **Factor the Equation**:
Factoring out \(t\):
\[ t(16t - 96) = 0 \]
This gives us two solutions:
\[ t = 0 \quad \text{or} \quad 16t - 96 = 0 \]

5. **Solve for t**:
Solving the second equation:
\[ 16t = 96 \implies t = \frac{96}{16} = 6 \text{ seconds} \]

6. **Conclusion**:
The body will reach the ground after **6 seconds**.