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.
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:
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
The position of the sandbag at 0.250 s after its release is 40.9 m
We want to know the height of the balloon as a function of time, after 0.250s
Then we use the following equation of motion:
In this equation
is the speed
is gravitational acceleration
is the initial height
is the initial velocity.
If the sandbag is released 40 meters above the ground, then the initial height is 40 meters.
As the balloon rises at a constant speed, then the initial velocity of the balloon is 5 m/s
Now we substitute these values in the formula
Now we look for:
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