Ahot-air balloonist, rising vertically with a constant speed of 5.00 m/s releases a sandbag at the instant the balloon is 40.0 m above the ground. after it is released, the sandbag encounters no appreciable air drag. compute the velocity of the sandbag at 0.250 s after its release.

 ↑   upwards

Explanation:

Take the direction up (where the hot-air balloonist moves) as the positive direction, and take the direction down (contrary to the direction of the hot-air balloonist) as the negative direction.

Then we propose the kinematic equation for the speed of the sandbag:

.

Where

is the initial velocity of the sandbag

a is the acceleration of the sandbag

t is the time in seconds.

In this case

a is the gravitational acceleration of

Now we substitute these values into the formula and solve for

 ↑   upwards

the components of a vector are the parts of the vectors that are to each other.

b perpendicular

the right answer is 0.3 meters per second