A person in a balloon moving vertically upward at a constant speed of 4.9 m/s drops a sandbag at an elevation of 98 m. What time will it take until the sandbag hits the ground?

To find the time it will take for the sandbag to hit the ground, we can use the kinematic equation:

s = ut + (1/2)at^2

Where:
s = distance (98 m)
u = initial velocity (0 m/s, as the sandbag is dropped)
a = acceleration due to gravity (-9.8 m/s^2, as it is acting downward)
t = time

Plugging in the values:

98 = 0*t + (1/2)(-9.8)t^2
98 = -4.9t^2
t^2 = 98 / 4.9
t^2 = 20
t = √20
t ≈ 4.47 seconds

Therefore, it will take approximately 4.47 seconds for the sandbag to hit the ground.

To find the time it takes for the sandbag to hit the ground, we need to consider its initial height, initial velocity (which is zero since it is dropped), and the acceleration due to gravity (g = 9.81 m/s²).

We can use the following equation to relate these quantities:

```
h = vi - t + 1/2 - g - t²
```

where:

- h is the initial height (98 m)
- vi is the initial velocity (0 m/s)
- t is the time (the unknown we want to find)
- g is the acceleration due to gravity (9.81 m/s²)

Since the sandbag is dropped from rest, the initial velocity (vi) is 0. Substituting the given values into the equation, we get:

```
98 = 0 - t + 1/2 - 9.81 - t²
```

Simplifying the equation:

```
98 = 4.905 - t²
```

Solving for t:

```
t² = 98 / 4.905
t² = 20
t = √20
t ≈ 4.47 s
```

Therefore, it will take approximately 4.47 seconds for the sandbag to hit the ground.