We can determine how long it will take for the watermelon to hit the ground by using the equation of motion:

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

Where:

• $\Delta d$ is the displacement (in this case, 5.20 m),
• $v_i$ is the initial velocity (which is 0 m/s because the watermelon is dropped from rest),
• $g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$),
• $t$ is the time we are solving for.

Since the watermelon is dropped from rest, $v_i = 0$, and the equation simplifies to:

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

Substituting the values:

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

Now we solve for $t$:

$5.20 = 4.905 t^2$

$t^2 = \frac{5.20}{4.905} \approx 1.06$

$t \approx \sqrt{1.06} \approx 1.03 \, \text{seconds}$

Thus, it should take approximately 1.03 seconds for the watermelon to hit the ground.

Would you like further details on any step or have questions? Here are 5 related questions:

1. How would the time change if the height was doubled?
2. How would air resistance affect this calculation?
3. What is the final velocity of the watermelon just before hitting the ground?
4. What would happen if the initial velocity was not zero?
5. How can you calculate the impact force on the watermelon?

Tip: The formula used here assumes no air resistance. In real-world conditions, factors like air drag might slightly alter the result.