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¢ + šš”
Where:
š£ is the final velocity,
š¢ is the initial velocity,
š 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) 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 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.