For this projectile motion problem, let's solve it step by step.

Given dataFor this projectile motion problem, let's solve it step by step.

Given

• The bullet is fired at an initial speed $v_0 = 427 \, \text{m/s}$.
• The horizontal distance to the target is $x = 91.4 \, \text{m}$.
• Acceleration due to gravity is $g = 9.8 \, \text{m/s}^2$.

Part (a): Finding the possible angles $\theta_1$ and $\theta_2$

Using the formula for the range of projectile motion: $R = \frac{v_0^2 \sin(2\theta)}{g}$ where:

• $R$ is the range or the horizontal distance,
• $v_0$ is the initial velocity,
• $\theta$ is the angle of projection,
• $g$ is the acceleration due to gravity.

Rearrange the equation to solve for $\theta$: $\sin(2\theta) = \frac{R \cdot g}{v_0^2}$ Substituting the given values: $\sin(2\theta) = \frac{91.4 \times 9.8}{427^2}$ Let me calculate this value.

We have: $\sin(2\theta) = 0.00491$

Now, to find the two possible angles, we solve for $2\theta$ using the inverse sine function: $2\theta = \sin^{-1}(0.00491)$ Then divide by 2 to get $\theta$.

Let me calculate the angles for you.The two possible angles are:

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

Part (b): Time of flight for both angles

To find the time of flight, we use the horizontal motion formula: $t = \frac{R}{v_0 \cos(\theta)}$ We will calculate the time for both $\theta_1$ and $\theta_2$. Let me proceed with that calculation.The time of flight for the two angles is:

• For $\theta_1 \approx 0.14^\circ$, the time is $t_1 \approx 0.214 \, \text{seconds}$.
• For $\theta_2 \approx 89.86^\circ$, the time is $t_2 \approx 87.14 \, \text{seconds}$.

These results show that the low-angle shot ($\theta_1$) results in a short time to reach the target, while the high-angle shot ($\theta_2$) takes a much longer time.

Would you like further details or clarifications? Here are some related questions:

1. What would happen to the angles if the speed of the bullet was doubled?
2. How does air resistance affect the actual time of flight in a real scenario?
3. What other factors could influence the accuracy of the shot over such a long distance?
4. Can the two angles be used in practical applications for shooting targets at different ranges?
5. How does gravity impact projectiles at different elevations?

Tip: For optimal shooting angles in real-life scenarios, using angles between 30° and 45° often results in the best balance between distance and speed.