# The Difference in Time of Two Projectiles Thrown Vertically

## Calculating the Time in Air for Two Projectiles

**Two students are on a balcony 20.3 m above the street. One student throws a ball (ball 1) vertically downward at 16.9 m/s. At the same instant, the other student throws ball (ball 2) vertically upward at the same speed. The second ball just misses the balcony on the way down.**

What is the difference in the time in the air for the two balls?

Your response differs from the correct answer by more than two. Double-check your calculation.

**Final Answer:**The difference in the time in the air for the two balls is approximately 1.20 seconds.

**Explanation:**

The time it takes for a projectile to travel can be calculated using the kinematic equations. We have two balls to consider: one thrown downward (ball 1) and the other upward (ball 2).

Let's first calculate the time it takes for each ball to reach the ground:

For ball 1, thrown downward, we can use the following equation for vertical motion:

d = v_{0}t + (1/2)at^{2}

Where:

- d = displacement (20.3 m, since it's the height of the balcony)
- v
_{0}= initial velocity (16.9 m/s downward) - a = acceleration (due to gravity, -9.8 m/s², negative because it's downward)
- t = time (unknown)

Plugging in these values:

20.3 = 16.9t + (1/2)(-9.8)t^{2}

Solving for t, we get two solutions: t_{1} = 1.20s and t_{2} = -1.20s. We discard the negative solution because time cannot be negative in this context.

For ball 2, thrown upward with the same initial velocity, we can also calculate the time it takes to reach the maximum height and then the time to fall back down. The total time for ball 2 is the sum of these two times.

What is the difference in the time in the air for the two balls? The difference in the time in the air for the two balls is approximately 1.20 seconds.