To determine when the ball hits the ground, we set the distance $D$ to 50 feet, because that is the height of the bridge.

The equation given is: $D = 16t^2$

1. Substitute $D = 50$ into the equation: $50 = 16t^2$

2. Solve for $t^2$ by dividing both sides by 16: $t^2 = \frac{50}{16} = 3.125$

3. Take the square root of both sides to solve for $t$: $t = \sqrt{3.125} \approx 1.7678$

4. Rounding to the nearest thousandth, we get: $t \approx 1.768 \text{ seconds}$

Answer:

The ball will hit the ground at approximately $t = 1.768$ seconds.

Would you like a breakdown of each calculation step?

Here are some related questions:

1. How would the time change if the height of the bridge were doubled?
2. What is the impact on time if gravity was weaker (e.g., on the Moon)?
3. How does the formula $D = 16t^2$ relate to the concept of gravitational acceleration?
4. What would be the formula if the initial velocity wasn’t zero?
5. How does air resistance affect the actual time in real-world conditions?

Tip: Remember that in physics problems involving free fall, assuming negligible air resistance simplifies calculations.