Let's break down the problem into parts and solve each step by step.

Part A: Position and velocity of the sandbag 2.00 seconds after its release

We are given:

• Initial height of the sandbag $h_0 = 50.0 \ \text{m}$
• Speed of the balloon $v_0 = 3.00 \ \text{m/s}$ (this is also the initial velocity of the sandbag because it was released from the balloon)
• Acceleration due to gravity $g = 9.81 \ \text{m/s}^2$ (acting downward)
• Time $t = 2.00 \ \text{s}$

We need to find the position and velocity of the sandbag after 2 seconds.

1. Velocity of the sandbag after 2.00 seconds:

The velocity at any time $t$ after release is given by the equation: $v(t) = v_0 - g t$ Substituting the known values: $v(2.00 \ \text{s}) = 3.00 \ \text{m/s} - 9.81 \ \text{m/s}^2 \times 2.00 \ \text{s}$

2. Position of the sandbag after 2.00 seconds:

The position (height) at any time $t$ is given by the kinematic equation: $y(t) = h_0 + v_0 t - \frac{1}{2} g t^2$ Substituting the known values: $y(2.00 \ \text{s}) = 50.0 \ \text{m} + 3.00 \ \text{m/s} \times 2.00 \ \text{s} - \frac{1}{2} \times 9.81 \ \text{m/s}^2 \times (2.00 \ \text{s})^2$

Let's calculate both the velocity and position.After 2.00 seconds:

• The velocity of the sandbag is $v = -16.62 \ \text{m/s}$ (downward).
• The position (height above the ground) is $y = 36.38 \ \text{m}$.

Part B: Time when the sandbag strikes the ground

We need to determine how long it takes for the sandbag to hit the ground. This happens when $y(t) = 0$.

Using the kinematic equation for position: $0 = h_0 + v_0 t - \frac{1}{2} g t^2$ This is a quadratic equation of the form: $0 = 50.0 + 3.00 t - 4.905 t^2$ We will solve this quadratic equation for $t$.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can still solve the quadratic equation manually or using a calculator:

The equation to solve is: $0 = -4.905 t^2 + 3.00 t + 50.0$

You can use the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Where:

• $a = -4.905$
• $b = 3.00$
• $c = 50.0$

Part C: Greatest height of the sandbag

The greatest height occurs when the velocity becomes zero (i.e., when it momentarily stops going up and starts to fall).

Use the equation for velocity: $v(t) = v_0 - g t = 0$ Solve for $t$ to find the time at which the sandbag reaches its maximum height, then substitute this value into the position equation.

If you need more details or help with the calculations, feel free to ask!

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Here are five related questions:

1. How would the solution change if the balloon's speed were different?
2. How does air resistance affect the motion of a falling object like the sandbag?
3. What happens to the sandbag's acceleration throughout its fall?
4. Can you derive the equations of motion used in this problem?
5. What would happen if the sandbag was thrown upwards instead of just released?

Tip:

When solving quadratic equations, always check for two possible solutions (positive and negative time) and select the one that makes physical sense (positive time).