To solve this problem, we need to calculate the time required for a bullet to reach a target located at a distance of 91.4 meters, given two different launch angles. The initial speed of the bullet ($v_0$) is 427 m/s.

Step-by-Step Solution:

1. Determine the Time of Flight Equation:

   The range $R$ of a projectile is given by:

   $R = \frac{v_0^2 \sin(2\theta)}{g}$

   where:

   • $R = 91.4$ m (range to the target)
   • $v_0 = 427$ m/s (initial velocity)
   • $g = 9.8$ m/s$^2$ (acceleration due to gravity)
   • $\theta$ is the launch angle.

   We need to solve for the two possible angles $\theta_1$ and $\theta_2$.

2. Solve for $\theta$:

   Rearrange the equation to solve for $\theta$:

   $\sin(2\theta) = \frac{Rg}{v_0^2}$

   Plugging in the values:

   $\sin(2\theta) = \frac{91.4 \times 9.8}{427^2}$

   Let's calculate the value of $\sin(2\theta)$.The value of $\sin(2\theta)$ is approximately $0.00491$.

3. Calculate the Two Possible Angles:

   Since $\sin(2\theta) = 0.00491$, we can find the two possible values of $2\theta$:

   $2\theta_1 = \arcsin(0.00491), \quad 2\theta_2 = 180^\circ - \arcsin(0.00491)$

   Now, let's compute the angles $\theta_1$ and $\theta_2$ by dividing these by 2.The two possible launch angles are:

• $\theta_1 \approx 0.14^\circ$
• $\theta_2 \approx 89.86^\circ$

4. Determine the Time of Flight for Each Angle:

The time of flight $t$ for a projectile is given by:

$t = \frac{2v_0 \sin(\theta)}{g}$

We will calculate the time required for both angles, $\theta_1$ and $\theta_2$.The time required for the bullet to reach the target for the two launch angles is:

• For $\theta_1 \approx 0.14^\circ$: $t_1 \approx 0.21$ seconds
• For $\theta_2 \approx 89.86^\circ$: $t_2 \approx 87.14$ seconds

Explanation

• At a very low angle ($\theta_1$), the bullet travels almost horizontally, reaching the target quickly.
• At a very high angle ($\theta_2$), the bullet travels nearly vertically before coming down, taking a significantly longer time to reach the same target.

Would you like further details or have any questions?

Follow-Up Questions:

1. How does changing the initial velocity affect the time of flight for both angles?
2. What happens to the range if the angle is increased beyond 45 degrees?
3. How would air resistance affect the calculated times for both angles?
4. Can you derive the maximum height reached by the bullet for both angles?
5. How do these angles relate to achieving maximum range for a given velocity?

Tip:

Always remember that for projectile motion, there are two angles that can achieve the same range but differ in their time of flight.