A hot-air balloon is ascending at the rate of 13 m/s and is 62 m above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?

hf=hi+Vi*t-1/2 g t^2

0=62+13t-4.8t^2 solve for t.

To solve this problem, we need to use the equations of motion, particularly those for vertical motion. The key equations for this situation are:

1. Displacement: s = ut + (1/2)at^2
2. Final velocity: v = u + at
3. Time: t = (v - u) / a

Where:
- s is the displacement (height in this case),
- u is the initial velocity,
- a is the acceleration, which in this case is due to gravity and is approximately 9.8 m/s^2,
- t is the time, and
- v is the final velocity.

(a) How long does the package take to reach the ground?
Since the package is dropped from rest, its initial velocity (u) is 0. The acceleration (a) is the acceleration due to gravity, which is -9.8 m/s^2 (negative because it is directed downward). The displacement (s) is the height of the balloon, which is 62 m.

Using the displacement equation, we can solve for time (t):
s = ut + (1/2)at^2
62 = 0*t + (1/2)*(-9.8)*t^2
62 = -4.9t^2

Rearranging the equation:
t^2 = 62 / -4.9

Taking the square root of both sides to determine the positive time:
t = √(62 / -4.9)

Calculating the time, we find:
t ≈ 3.17 seconds

So, the package takes approximately 3.17 seconds to reach the ground.

(b) With what speed does it hit the ground?
To find the final velocity (v), we can use the final velocity equation:
v = u + at

Since the package is dropped, the initial velocity (u) is 0, and the acceleration (a) is -9.8 m/s^2. We can substitute these values into the equation:
v = 0 + (-9.8) * 3.17

Calculating the final velocity, we get:
v ≈ -31.07 m/s

The negative sign indicates that the velocity is directed downward. However, we're only concerned with the magnitude of the velocity, so the speed at which the package hits the ground is approximately 31.07 m/s.