To solve this problem, we need to find the launch speed of a projectile given its horizontal distance, time of flight, and launch angle.

Given Data:

• Horizontal distance ($R$): $200 \, \text{m}$
• Time of flight ($t$): $5.0 \, \text{s}$
• Launch angle ($\theta$): $45^\circ$

Solution:

The horizontal distance for a projectile is given by: $R = v_x \cdot t$ where $v_x$ is the horizontal component of the initial velocity. Since the launch angle is $45^\circ$, the horizontal and vertical components of the initial velocity ($v_0$) are equal.

Step 1: Find the horizontal component $v_x$:

The horizontal component of the initial velocity is: $v_x = v_0 \cdot \cos(45^\circ)$ And since $\cos(45^\circ) = \frac{\sqrt{2}}{2}$, we can substitute and rearrange the formula for $R$: $R = v_0 \cdot \cos(45^\circ) \cdot t$ $200 = v_0 \cdot \frac{\sqrt{2}}{2} \cdot 5$ $v_0 = \frac{200}{\frac{\sqrt{2}}{2} \cdot 5} = \frac{200 \times 2}{5 \cdot \sqrt{2}} = \frac{400}{5 \sqrt{2}} = \frac{80}{\sqrt{2}}$ $v_0 = 80 \times \frac{\sqrt{2}}{2} = 80 \times 0.7071 \approx 56.6 \, \text{m/s}$

Answer:

The closest answer choice is: A) 56 m/s

Would you like further details or have any questions?

Here are five related questions to expand your understanding:

1. How would the launch speed change if the angle were 30° instead of 45°?
2. What is the maximum height achieved by the projectile in this scenario?
3. How would you calculate the horizontal range if only the initial speed and launch angle were given?
4. What factors affect the time of flight of a projectile?
5. How would the answer change if air resistance were considered?

Tip: Remember, for a projectile launched at 45°, the horizontal and vertical components of velocity are equal, simplifying calculations when using symmetry in motion.