A hot air balloon is rising upwards at a constant velocity of 5m.s,when the balloon is 60mabove the ground ,a sandbag is released from it and allowed to free fall ,ignore the effects of air
1 describe the motion of sandbag when it was dropped free fall
2 where is the sandbag after 3seconds
1. When the sandbag is dropped from the hot air balloon, it will experience free fall. In free fall, the object falls under the influence of gravity alone, without any other forces acting on it. Since we are ignoring the effects of air, the sandbag will accelerate downwards at a constant rate of 9.8 m/s² due to gravity. This means that its velocity will continuously increase at a rate of 9.8 m/s per second.
2. To determine the position of the sandbag after 3 seconds, we can use the equations of motion for an object in free fall. In this case, we need to consider that the initial velocity of the sandbag is 0 m/s because it was dropped from rest.
The equation to calculate the position of an object in free fall is given by:
d = (1/2)gt²
where d is the distance traveled, g is the acceleration due to gravity (9.8 m/s²), and t is the time in seconds.
Substituting the values into the equation, we get:
d = (1/2)(9.8 m/s²)(3 s)²
= (1/2)(9.8 m/s²)(9 s²)
= 44.1 m
Therefore, after 3 seconds, the sandbag will be approximately 44.1 meters below the point it was released from.