A projectile has a maximum range of 16 km. At the highest point of its motion, it explodes into two equal masses. One mass drops vertically downward. The horizontal distance covered by the other mass from the time of explosion

Options

A. 8 km
B. 16 km
C. 24 km
D. 32 km

Answer

To solve the problem step by step, we can follow these logical steps:

### Step 1: Understand the problem
We have a projectile that reaches a maximum range of 16 km. At its highest point, it explodes into two equal masses. One mass drops vertically, and we need to find the horizontal distance covered by the other mass from the time of explosion.

### Step 2: Determine the initial conditions
The maximum range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

Since the maximum range occurs at \( \theta = 45^\circ \), we can substitute \( \sin 90^\circ = 1 \):
\[ R = \frac{u^2}{g} \]
Given that \( R = 16 \text{ km} = 16000 \text{ m} \), we have:
\[ \frac{u^2}{g} = 16000 \]

### Step 3: Analyze the explosion
At the highest point of the projectile's motion, the horizontal component of the velocity \( u_x \) is given by:
\[ u_x = u \cos \theta \]
For \( \theta = 45^\circ \):
\[ u_x = u \cdot \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}} \]

### Step 4: Determine the time of flight after the explosion
The time taken to reach the maximum height (where the explosion occurs) can be calculated using:
\[ t = \frac{u \sin \theta}{g} \]
Since the projectile travels half the range to reach the maximum height:
\[ t = \frac{u \cdot \frac{1}{\sqrt{2}}}{g} = \frac{u}{\sqrt{2}g} \]
The total time of flight for the projectile is \( 2t \), but since we only need the time after the explosion to reach the ground, we can use:
\[ t_{fall} = \frac{u \sin \theta}{g} = \frac{u}{\sqrt{2}g} \]

### Step 5: Calculate the horizontal distance covered by the second mass
The horizontal distance \( d \) covered by the mass that continues to move horizontally after the explosion can be calculated as:
\[ d = u_x \cdot t_{fall} \]
Substituting the values we have:
\[ d = \left(\frac{u}{\sqrt{2}}\right) \cdot \left(\frac{u}{\sqrt{2}g}\right) = \frac{u^2}{2g} \]
From Step 2, we know \( \frac{u^2}{g} = 16000 \):
\[ d = \frac{16000}{2} = 8000 \text{ m} = 8 \text{ km} \]

### Step 6: Conclusion
The horizontal distance covered by the mass after the explosion is:
\[ \boxed{8 \text{ km}} \]