A hot air balloon is traveling vertically upward at a constant speed of 3.9 m/s. When it is 20 m above the ground, a package is released from the balloon. After it is released, for how long is the package in the air? The acceleration of gravity is 9.8 m/s
A HOT AIR BALLOON IS RISING UPWARDS AT A CONSTANT VELOCITY OF 5 M/S. WHEN THE BALLOON IS 100M ABOVE THE GROUND, A SAND BAG IS DROPPED FROM IT.WHAT IS THE MAGNITUDE OF THE ACCELERATION OF THE HOT AIR BALLOON WHILE THE SANDBAG IS IN IT
To find the time the package is in the air after it is released from the balloon, we can use the equation of motion for vertical motion:
h = ut + (1/2)at^2
Where:
h = height (distance)
u = initial velocity
a = acceleration
t = time
In this case, the initial height of the package above the ground is 20 m, the initial velocity is the same as the balloon's upward velocity (3.9 m/s), the acceleration is the acceleration due to gravity (-9.8 m/s^2), and we want to find the time (t).
Rearranging the equation, we have:
t = √((2h) / a)
Substituting the given values:
t = √((2 * 20) / -9.8)
t = √(-40 / -9.8)
t = √4.0816
t ≈ 2.02 seconds
Therefore, the package is in the air for approximately 2.02 seconds after it is released from the balloon.