A hotair balloonist, rising vertically with a constant speed of 5.00 m/s, releases a sadndbag at the instant the balloon is 40.0 m above the ground. after it is released, the sandbag encounters no appreciable air drag. a. compute the position and velocity of the sandbag at .250 s and 1.00 s after its release. b. how many seconds after its release will the bag strike the ground? C. how fast is it moving as it strikes the ground? d. what is the greatest height above the ground that the sandbag reaches?

a. To compute the position and velocity of the sandbag at different times, we can use the equations for uniformly accelerated motion.

Given:
Initial position (h₀) = 40.0 m
Initial velocity (v₀) = 0 m/s (The sandbag is initially at rest)
Acceleration (a) = gravitational acceleration = 9.8 m/s² (downwards direction)

Using the equations:
Position (h) = h₀ + v₀t + (1/2)at²
Velocity (v) = v₀ + at

At t = 0.250 s:
Position (h) = 40.0 m + (0 m/s)(0.250 s) + (1/2)(9.8 m/s²)(0.250 s)²
Position (h) = 40.0 m + 0 m + (1/2)(9.8 m/s²)(0.0625 s²)
Position (h) = 40.0 m + 0 m + 0.30625 m
Position (h) = 40.30625 m

Velocity (v) = 0 m/s + (9.8 m/s²)(0.250 s)
Velocity (v) = 0 m/s + 2.45 m/s
Velocity (v) = 2.45 m/s

At t = 1.00 s:
Position (h) = 40.0 m + (0 m/s)(1.00 s) + (1/2)(9.8 m/s²)(1.00 s)²
Position (h) = 40.0 m + 0 m + (1/2)(9.8 m/s²)(1.00 s)²
Position (h) = 40.0 m + 0 m + 4.9 m
Position (h) = 44.9 m

Velocity (v) = 0 m/s + (9.8 m/s²)(1.00 s)
Velocity (v) = 0 m/s + 9.8 m/s
Velocity (v) = 9.8 m/s

b. To find the time it takes for the sandbag to strike the ground, we need to solve for t when h = 0.

Using the equation:
h = h₀ + v₀t + (1/2)at²

0 = 40.0 m + 0 m/s * t + (1/2)(9.8 m/s²)t²

This is a quadratic equation, and we can solve it for t using the quadratic formula or by factoring. Let's use the quadratic formula:

t = (-b ± √(b² - 4ac)) / (2a)

In this case, a = (1/2)(-9.8 m/s²), b = 0 m/s, and c = 40.0 m.

t = (-0 ± √(0² - 4(1/2)(-9.8 m/s²)(40.0 m))) / (2(1/2)(-9.8 m/s²))
t = √(4(1/2)(9.8 m/s²)(40.0 m)) / (9.8 m/s²)
t = √(2(9.8 m/s²)(40.0 m)) / (9.8 m/s²)
t = √((2 * 9.8 m/s² * 40.0 m) / 9.8 m/s²)
t = √(78.4 s²)
t ≈ 8.84 s

Therefore, the sandbag will strike the ground approximately 8.84 seconds after its release.

c. To find the speed of the sandbag as it strikes the ground, we can use the equation for velocity:

Velocity (v) = v₀ + at

Since the sandbag is initially at rest (v₀ = 0 m/s), we have:

Velocity (v) = 0 m/s + (9.8 m/s²)(8.84 s)
Velocity (v) ≈ 86.55 m/s

Therefore, the sandbag is moving with a velocity of approximately 86.55 m/s as it strikes the ground.

d. To find the maximum height above the ground that the sandbag reaches, we can use the equation for the position at any time:

Position (h) = h₀ + v₀t + (1/2)at²

The maximum height occurs when v = 0 m/s. Let's set v = 0 and solve for t:

0 = 0 m/s + (9.8 m/s²)t
t = 0 s

This means that at t = 0 s, the sandbag is at its highest point. Therefore, the maximum height above the ground that the sandbag reaches is 40.0 m.

To compute the position and velocity of the sandbag at different times after its release, we can use the equations of motion for constant acceleration. In this case, the acceleration is due to gravity, which is approximately 9.8 m/s² downward.

a. At time t = 0.250 s after its release:
- Position: The initial position is 40.0 m above the ground. Since the sandbag is in free fall, the equation for position is given by: s = ut + (1/2)at², where s is the position, u is the initial velocity, t is the time, and a is the acceleration.
- Plug in the values: s = 40.0 m + (0 m/s)(0.250 s) + (1/2)(-9.8 m/s²)(0.250 s)² = 40.0 m - (0.30625) m = 39.7 m (approximately)

- Velocity: The velocity remains constant at 5.00 m/s, as there is no air drag.
- The velocity remains 5.00 m/s.

At time t = 1.00 s after its release:
- Position: Using the same equation as above: s = 40.0 m + (0 m/s)(1.00 s) + (1/2)(-9.8 m/s²)(1.00 s)² = 39.0 m (approximately)

- Velocity: The velocity remains constant at 5.00 m/s.
- The velocity remains 5.00 m/s.

b. To calculate the time it takes for the sandbag to strike the ground, we can use the equation for position and solve for time when the position is equal to 0. Since the sandbag is in free fall, the equation for position is: s = ut + (1/2)at².
- Set s = 0 and solve for t: 0 = 40.0 m + (0 m/s)t + (1/2)(-9.8 m/s²)t²
- This is a quadratic equation, and solving it will give two possible values for time, one positive and one negative. We take the positive value since time cannot be negative.
- Using the quadratic formula, t = (-b ± √(b² - 4ac)) / (2a), where a = (1/2)(-9.8 m/s²), b = 0 m/s, and c = 40.0 m.
- Plug in the values and solve: t = (-0 ± √(0² - 4(1/2)(-9.8)(40))) / (2(1/2)(-9.8)) = 2.02 s (approximately)
- Therefore, it takes approximately 2.02 seconds for the sandbag to strike the ground.

c. To determine the speed of the sandbag as it strikes the ground, we can use the equation for velocity: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
- Plug in the values: v = 0 m/s + (-9.8 m/s²)(2.02 s) = approximately -19.8 m/s
- The magnitude of the velocity is taken, so the speed of the sandbag as it strikes the ground is approximately 19.8 m/s.

d. To find the greatest height reached by the sandbag, we can consider the time when its velocity becomes zero. At this point, the sandbag will start falling back down.
- Use the equation for velocity: v = u + at.
- Set v = 0, u = 5.00 m/s, and a = 9.8 m/s² (taking into account the acceleration due to gravity).
- Solve for t: 0 = 5.00 m/s + 9.8 m/s²t.
- t = - (5.00 / 9.8) ≈ -0.51 s (approximately).
- Since time cannot be negative, we discard this value.
- Therefore, the sandbag does not reach any height greater than its initial height above the ground.

In summary:
a) At t = 0.250 s, the position is approximately 39.7 m and the velocity is 5.00 m/s.
At t = 1.00 s, the position is approximately 39.0 m and the velocity is 5.00 m/s.
b) The sandbag strikes the ground after approximately 2.02 seconds.
c) The speed of the sandbag as it strikes the ground is approximately 19.8 m/s.
d) The sandbag does not reach any height greater than its initial height above the ground (40.0 m).

Vi=0 in the Y-direction of the bag going down, a=g=9.8m/s^2, d=40m. Even if it is rising at 5m/s, the velocity going down will not no be affected by the velocity going up. Since the velocity is 0, initially, and it accelerates down at a rate of 9.8m/s^2, multiply 9.8m/s^2 by 0.250s to give you the velocity at this time interval; notice that one of the s's cancel to give you units of m/s which is velocity at that time. In one second, the sand bag accelerated by 9.8m/s^2. Since the initial velocity (Vi) is 0 at the time of release, in one second the velocity is equal to gravity=9.8m/s. Use the formula (Vf)^2=(Vi)^2 +2ad to find the final velocity (Vf). Since Vi=0, the formula turns into Vf=(2ad)^1/2=(2(9.8m/s^2)(40m))^1/2. Once you compute the Vf, plug in Vf, Vi=0, and a=9.8m/s^2 into one of your other kinematic equations (Vf=Vi+at) and solve for time (t). Since the bag is released at 40m, it only reaches a max height of 40m.