A small mailbag is released from a helicopter that is descending steadily at 2.87 m/s.
a)After 5.00 s, what is the speed of the mailbag?
b)How far is it below the helicopter?
a. vf=vi+gt
g is downward, as is vi
b. distance= vi*t+1/2 g t^2 again, vi, g are negative.
To answer these questions, we need to consider the motion of the mailbag as it is released from the descending helicopter.
a) After 5.00 s, what is the speed of the mailbag?
To find the speed of the mailbag after 5.00 s, we can use the formula for velocity:
velocity = initial velocity + acceleration × time
In this case, the initial velocity of the mailbag is the velocity of the helicopter, which is descending steadily at 2.87 m/s. The acceleration of the mailbag can be considered zero since there is no horizontal force acting on it (assuming air resistance is negligible). Therefore, the equation simplifies to:
velocity = initial velocity
So, after 5.00 s, the speed of the mailbag will also be 2.87 m/s.
b) How far is it below the helicopter?
To find the distance traveled by the mailbag below the helicopter, we can use the formula for displacement:
displacement = initial velocity × time + (1/2) × acceleration × time^2
Again, in this case, the initial velocity and acceleration of the mailbag are both zero since there is no horizontal force acting on it. Therefore, the equation simplifies to:
displacement = 0 × time + (1/2) × 0 × time^2
displacement = 0
So, the mailbag will be at a distance of 0 meters below the helicopter after 5.00 s.