what is the velocity (in ft/s) of a sandbag 2.75 s after it is released from a hot-air balloon that is stationary in the air?

To find the velocity of a sandbag 2.75 seconds after it is released from a stationary hot-air balloon, we need to use the equation of motion known as the kinematic equation.

The specific kinematic equation required for this problem is:

š‘£ = š‘¢ + š‘Žš‘”

Where:
š‘£ is the final velocity,
š‘¢ is the initial velocity,
š‘Ž is the acceleration, and
š‘” is the time.

In this case, since the hot-air balloon is stationary, the initial velocity (š‘¢) is zero.

We also need to know the acceleration (š‘Ž) acting on the sandbag. Since the sandbag is free-falling under the influence of gravity, we can use the acceleration due to gravity, which is approximately 32.2 ft/sĀ².

Now, let's put the given values into the equation:

š‘£ = 0 + (32.2 ft/sĀ²)(2.75 s)

Calculating this, we get:

š‘£ ā‰ˆ 88.55 ft/s

Therefore, the velocity of the sandbag 2.75 seconds after it is released is approximately 88.55 ft/s.

V = Vo + g*t

Vo = 0
g = 9.8
t = 2.75 s.
Solve for V.