To solve this problem, we can use the kinematic equation provided:

$\Delta d = v_i t + \frac{1}{2} g t^2$

Here:

• $\Delta d$ is the displacement (5.20 m, the height of the bleachers),
• $v_i$ is the initial velocity (since the watermelon is dropped, $v_i = 0$),
• $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$),
• $t$ is the time we want to find.

Since the initial velocity $v_i = 0$, the equation simplifies to:

$\Delta d = \frac{1}{2} g t^2$

Substitute the known values:

$5.20 = \frac{1}{2} (9.8) t^2$

Simplify the equation:

$5.20 = 4.9 t^2$

Solve for $t^2$:

$t^2 = \frac{5.20}{4.9} = 1.0612$

Now, take the square root of both sides to solve for $t$:

$t = \sqrt{1.0612} \approx 1.03 \, \text{seconds}$

Thus, the time it takes for the watermelon to hit the ground is approximately 1.03 seconds, so the correct answer is:

A. 1.03 s

---

Would you like more details or have any questions? Here are some related questions to expand on this:

1. How would the time change if the height of the bleachers doubled?
2. What would happen if the watermelon were thrown downwards with an initial velocity?
3. How does the acceleration due to gravity influence the motion of falling objects?
4. If air resistance were significant, how would that affect the result?
5. What is the difference between free fall and projectile motion?

Tip: In free fall, the only force acting on the object is gravity, and the initial velocity is often zero unless the object is thrown.