A body is thrown horizontally from the top of a tower. It reaches the ground after 4s at an angle 45° to the ground. The velocity of projection is

Options

A. `9.8 ms^(-1)`
B. `19.6 ms^(-1)`
C. `29.4 ms^(-1)`
D. `39.2 ms^(-1)`

Answer

To solve the problem step by step, we will break down the information given and apply the relevant physics concepts.

### Step 1: Understand the problem
A body is thrown horizontally from the top of a tower and reaches the ground after 4 seconds at an angle of 45° to the ground. We need to find the initial velocity of projection.

### Step 2: Identify the time of flight
The time of flight (t) is given as 4 seconds.

### Step 3: Relate time of flight to height
The time of flight for a projectile thrown horizontally can be expressed using the formula:
\[ t = \sqrt{\frac{2h}{g}} \]
where:
- \( h \) is the height of the tower,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step 4: Solve for height (h)
Rearranging the formula to solve for height \( h \):
\[ h = \frac{1}{2} g t^2 \]
Substituting the values:
\[ h = \frac{1}{2} \times 9.8 \times (4^2) \]
\[ h = \frac{1}{2} \times 9.8 \times 16 \]
\[ h = 78.4 \, \text{m} \]

### Step 5: Analyze the angle of projection
The body hits the ground at an angle of 45°. This means the horizontal and vertical components of the velocity at the time of impact are equal.

### Step 6: Find the vertical velocity (Vy)
The vertical velocity \( V_y \) at the time of impact can be calculated using:
\[ V_y = g \cdot t \]
Substituting the values:
\[ V_y = 9.8 \cdot 4 \]
\[ V_y = 39.2 \, \text{m/s} \]

### Step 7: Relate horizontal and vertical velocities
Since the angle of impact is 45°, we have:
\[ \tan(45°) = \frac{V_y}{V_x} \]
Since \( \tan(45°) = 1 \):
\[ V_y = V_x \]
Thus,
\[ V_x = 39.2 \, \text{m/s} \]

### Step 8: Conclusion
The initial velocity of projection \( u \) is equal to the horizontal component \( V_x \):
\[ u = V_x = 39.2 \, \text{m/s} \]

### Final Answer
The velocity of projection is \( 39.2 \, \text{m/s} \).

---