Final answer:
After 1.00 s, the sandbag released from a rising hot-air balloon is 40.095 m above the ground and has a downward velocity of 4.81 m/s.
Step-by-step explanation:
Computing Position and Velocity of a Sandbag in Free Fall
The student is asking for the position and velocity of a sandbag 1.00 s after it has been released from a hot-air balloon rising vertically with a constant velocity. To calculate this, we'll use the principles of kinematics in physics. Since the balloon is moving upward at 5.00 m/s when the sandbag is released, the initial velocity (u) of the sandbag is 5.00 m/s upward. The acceleration (a) due to gravity is -9.81 m/s2, which will pull the sandbag downward.
Position after 1.00 s: We'll use the equation of motion s = ut + (1/2)at2, where s is the displacement. Substituting the values in, s = (5.00 m/s)(1 s) + (1/2)(-9.81 m/s2)(1 s)2, the displacement s = 5.00 m - 4.905 m = 0.095 m. Thus, the sandbag is 0.095 m above the release point, making it 40.095 m above the ground after 1 s.
Velocity after 1.00 s: We use the equation v = u + at, where v is the final velocity. Plugging in the values, v = 5.00 m/s + (-9.81 m/s2)(1 s) = -4.81 m/s. Therefore, the velocity of the sandbag after 1.00 s is 4.81 m/s downward.