A projectile is fired horizontally from a tower which is 25 ft high from the level of the ground and reaches the ground at a horizontal distance of 1000 ft. The initial velocity is

Options

A. `200 ft//"sec"`
B. `400 ft//"sec"`
C. `600 ft//"sec"`
D. `800 ft//"sec"`

Answer

To solve the problem of a projectile fired horizontally from a tower, we can break it down into a few steps:

### Step 1: Understand the Problem
We have a projectile fired horizontally from a height of 25 feet, and it lands 1000 feet away from the base of the tower. We need to find the initial horizontal velocity of the projectile.

### Step 2: Determine the Time of Flight
The time it takes for the projectile to fall can be calculated using the formula for vertical motion under gravity:
\[ h = \frac{1}{2} g t^2 \]
Where:
- \( h = 25 \) feet (height of the tower)
- \( g = 32 \) feet/s² (acceleration due to gravity)
- \( t \) = time in seconds

Substituting the values into the equation:
\[ 25 = \frac{1}{2} \times 32 \times t^2 \]
\[ 25 = 16 t^2 \]
\[ t^2 = \frac{25}{16} \]
\[ t = \sqrt{\frac{25}{16}} = \frac{5}{4} \text{ seconds} \]

### Step 3: Calculate the Initial Horizontal Velocity
The horizontal distance traveled by the projectile can be calculated using the formula:
\[ d = u \cdot t \]
Where:
- \( d = 1000 \) feet (horizontal distance)
- \( u \) = initial horizontal velocity (what we need to find)
- \( t = \frac{5}{4} \) seconds (time of flight)

Rearranging the formula to find \( u \):
\[ u = \frac{d}{t} \]
Substituting the known values:
\[ u = \frac{1000}{\frac{5}{4}} = 1000 \times \frac{4}{5} = 800 \text{ feet/second} \]

### Conclusion
The initial velocity of the projectile is **800 feet/second**.

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