Sure, let's break down the solution to the problem step by step. We're given that a bomb is dropped from a balloon and it reaches the ground in 30 seconds. We need to determine the height of the balloon in two scenarios: (a) when the balloon is at rest in the air, and (b) when it is ascending with a speed of 100 cm/s (which is equivalent to 1 m/s).
### Part a: Balloon at Rest
1. Time of Free Fall (t): 30 seconds.
2. Acceleration due to Gravity (g): 9.8 m/s².
3. Initial Velocity (u): Since the balloon is at rest, [tex]\( u = 0 \)[/tex] m/s.
We can use the equation of motion for distance under constant acceleration:
[tex]\[ s = ut + \frac{1}{2}gt^2 \][/tex]
Substituting the known values:
[tex]\[ s = (0 \times 30) + \frac{1}{2} \times 9.8 \times (30)^2 \][/tex]
[tex]\[ s = 0 + 0.5 \times 9.8 \times 900 \][/tex]
[tex]\[ s = 0.5 \times 9.8 \times 900 \][/tex]
[tex]\[ s = 4.9 \times 900 \][/tex]
[tex]\[ s = 4410 \][/tex]
So, the height of the balloon when at rest is [tex]\( \mathbf{4410} \)[/tex] meters.
### Part b: Balloon Ascending
1. Time of Free Fall (t): 30 seconds.
2. Acceleration due to Gravity (g): 9.8 m/s².
3. Ascending Speed (initial velocity), [tex]\( u \)[/tex]: 100 cm/s, converted to m/s is 1 m/s.
Again, use the equation of motion for distance:
[tex]\[ s = ut + \frac{1}{2}gt^2 \][/tex]
Substituting the known values:
[tex]\[ s = (1 \times 30) + \frac{1}{2} \times 9.8 \times (30)^2 \][/tex]
[tex]\[ s = 30 + 0.5 \times 9.8 \times 900 \][/tex]
[tex]\[ s = 30 + 0.5 \times 9.8 \times 900 \][/tex]
[tex]\[ s = 30 + 4.9 \times 900 \][/tex]
[tex]\[ s = 30 + 4410 \][/tex]
[tex]\[ s = 4410 + 30 \][/tex]
[tex]\[ s = 5310 \][/tex]
So, the height of the balloon when it is ascending with a speed of 1 m/s is [tex]\( \mathbf{5310} \)[/tex] meters.
### Summary:
- Height of the balloon when at rest: 4410 meters
- Height of the balloon when ascending at 1 m/s: 5310 meters
