A shell bursts on contact with the gorund and pieces from it fly in all directions with velocities up to `60 m//s`. Show that a man `180 m` away is in danger for `6sqrt(2) s`.

Answer

To solve the problem step by step, we will use the concepts of projectile motion.

### Step 1: Understanding the Problem
A shell bursts and pieces fly in all directions with a maximum velocity of 60 m/s. We need to determine how long a man standing 180 m away is in danger from these pieces.

### Step 2: Determine the Range
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where:
- \( u = 60 \, \text{m/s} \) (initial velocity),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( \theta \) is the angle of projection.

Since the man is 180 m away, we set \( R = 180 \, \text{m} \):
\[ 180 = \frac{60^2 \sin(2\theta)}{10} \]

### Step 3: Solve for \( \sin(2\theta) \)
Rearranging the equation gives:
\[ \sin(2\theta) = \frac{180 \times 10}{3600} = \frac{1800}{3600} = \frac{1}{2} \]

### Step 4: Find Possible Angles
The angles \( 2\theta \) that satisfy \( \sin(2\theta) = \frac{1}{2} \) are:
\[ 2\theta = 30^\circ \quad \text{or} \quad 2\theta = 150^\circ \]
Thus, the possible values for \( \theta \) are:
\[ \theta = 15^\circ \quad \text{or} \quad \theta = 75^\circ \]

### Step 5: Calculate Time of Flight for Each Angle
The time of flight \( T \) for a projectile is given by:
\[ T = \frac{2u \sin(\theta)}{g} \]

**For \( \theta = 15^\circ \):**
\[ T_1 = \frac{2 \times 60 \times \sin(15^\circ)}{10} \]
Using \( \sin(15^\circ) = \frac{\sqrt{6} - \sqrt{2}}{4} \):
\[ T_1 = \frac{120 \times \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)}{10} = \frac{12(\sqrt{6} - \sqrt{2})}{2} = 6(\sqrt{6} - \sqrt{2}) \]

**For \( \theta = 75^\circ \):**
\[ T_2 = \frac{2 \times 60 \times \sin(75^\circ)}{10} \]
Using \( \sin(75^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4} \):
\[ T_2 = \frac{120 \times \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)}{10} = \frac{12(\sqrt{6} + \sqrt{2})}{2} = 6(\sqrt{6} + \sqrt{2}) \]

### Step 6: Calculate the Danger Period
The danger period \( T_d \) is the difference between the two times of flight:
\[ T_d = T_2 - T_1 = 6(\sqrt{6} + \sqrt{2}) - 6(\sqrt{6} - \sqrt{2}) = 6\sqrt{2} + 6\sqrt{6} - 6\sqrt{6} + 6\sqrt{2} = 12\sqrt{2} \]

### Conclusion
The man is in danger for \( 6\sqrt{2} \, \text{s} \).

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