Jai is standing on the top of a building of height 25 m he wants to throw his gun to Veeru who stands on top of another building of height 20 m at distance 15 m from first building. For which horizontal speed of projectile, it is possible?

Options

A. `5 ms^(-1)`
B. `10 ms^(-1)`
C. `15 ms^(-1)`
D. `20 ms^(-1)`

Explanation

To solve the problem step by step, we need to analyze the motion of the projectile (the gun) that Jai throws from the top of a building to Veeru on another building.

### Step 1: Understand the vertical and horizontal distances
- Jai is on a building of height 25 m.
- Veeru is on a building of height 20 m.
- The vertical distance the gun needs to fall is:
\[ \text{Vertical distance} = 25 \, \text{m} - 20 \, \text{m} = 5 \, \text{m} \]
- The horizontal distance between the two buildings is 15 m.

### Step 2: Use the equations of motion for vertical motion
- The vertical motion can be described by the second equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
where:
- \( s \) is the vertical distance (5 m),
- \( u \) is the initial vertical velocity (0 m/s, since it is thrown horizontally),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( t \) is the time in seconds.

Substituting the known values into the equation:
\[ 5 = 0 \cdot t + \frac{1}{2} \cdot 9.81 \cdot t^2 \]
This simplifies to:
\[ 5 = \frac{1}{2} \cdot 9.81 \cdot t^2 \]
\[ 5 = 4.905 \cdot t^2 \]

### Step 3: Solve for time \( t \)
Rearranging the equation to solve for \( t^2 \):
\[ t^2 = \frac{5}{4.905} \]
Calculating \( t^2 \):
\[ t^2 \approx 1.0204 \]
Taking the square root to find \( t \):
\[ t \approx 1.01 \, \text{s} \quad (\text{approximately } 1 \, \text{s}) \]

### Step 4: Use the horizontal motion to find the required speed
- The horizontal motion can be described by the equation:
\[ x = ut \]
where:
- \( x \) is the horizontal distance (15 m),
- \( u \) is the horizontal speed (what we need to find),
- \( t \) is the time (approximately 1 s).

Substituting the known values:
\[ 15 = u \cdot 1 \]
Thus, solving for \( u \):
\[ u = 15 \, \text{m/s} \]

### Final Answer
The horizontal speed of the projectile required for Jai to successfully throw the gun to Veeru is **15 m/s**.

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