We can solve this problem using the given kinematic equation:

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

Where:

• $\Delta d$ is the displacement (5.20 m, the height from which the watermelon is dropped),
• $v_i$ is the initial velocity (since the watermelon is dropped, $v_i = 0$),
• $g$ is the acceleration due to gravity ($9.81 \, \text{m/s}^2$),
• $t$ is the time it takes to hit the ground.

Step 1: Simplify the equation

Since the initial velocity is 0, the equation becomes:

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

Substitute the known values:

$5.20 = \frac{1}{2} (9.81) t^2$

Step 2: Solve for $t^2$

$5.20 = 4.905 t^2$

Now, divide both sides by 4.905:

$t^2 = \frac{5.20}{4.905}$

$t^2 \approx 1.06$

Step 3: Solve for $t$

$t \approx \sqrt{1.06}$

$t \approx 1.03 \, \text{seconds}$

Conclusion:

The watermelon should take approximately 1.03 seconds to hit the ground.

Would you like more details or have any questions?

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Here are 5 related questions to explore:

1. What would happen if the initial velocity wasn't zero?
2. How would the time change if the bleachers were twice as high?
3. What is the final velocity of the watermelon when it hits the ground?
4. How does air resistance affect this scenario in real life?
5. What if the watermelon was thrown upwards first?

Tip: Always double-check units in physics problems to ensure accuracy in calculations.