Question

A hot-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.

Answer 1

Answer:

$v = 2.55\ m / s$  ↑   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:

$v (t) = v_0 + at$.

Where

$v_0$ 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 $9.8\ m / s ^ 2$

$v_0 = 5\ m / s$

$t = 0.250\ s$

Now we substitute these values into the formula and solve for $v(t)$

$v = 5 + (-9.8) (0.250)\\\\v = 5 -9.8 (0.250)$

$v = 2.55\ m / s$  ↑   upwards