Final answer:
The position and velocity of the sandbag at 0.250 s and 1.00 s after release can be calculated using kinematic equations. At 0.250 s, the sandbag is at 40.71 m with a velocity of 2.55 m/s upward. At 1.00 s, the position is 35.10 m with a velocity of -4.80 m/s downward.
Step-by-step explanation:
The question relates to the motion of a projectile in Physics, more specifically, involving a hot-air balloonist releasing a sandbag while ascending with constant velocity. To calculate the position and velocity of the sandbag at different time intervals after release, principles of kinematics are applied.
Assuming an upward direction is positive and acceleration due to gravity (g) is -9.8 m/s2, the initial velocity of the sandbag when released is +5.00 m/s. The position (y) of the sandbag after t seconds can be found using the formula:
y = y0 + v0t - ½gt2
where y0 is the initial height (40.0 m), v0 is the initial velocity, g is the acceleration due to gravity, and t is the time elapsed.
The velocity (v) of the sandbag at time t can be calculated using:
v = v0 - gt.
At 0.250 s after release:
Position: y = 40.0 m + (5.00 m/s)(0.250 s) - ½(9.8 m/s2)(0.250 s)2 = 40.71 m
Velocity: v = 5.00 m/s - (9.8 m/s2)(0.250 s) = 2.55 m/s.
At 1.00 s after release:
Position: y = 40.0 m + (5.00 m/s)(1.00 s) - ½(9.8 m/s2)(1.00 s)2 = 35.10 m
Velocity: v = 5.00 m/s - (9.8 m/s2)(1.00 s) = -4.80 m/s(downward).