A hot-air balloon is rising upward with a constant speed of 2.50 m/s. When the balloon is 3.00 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?
Vf=vi+gt -8,07=2,5+(-9,8)t t=0,57s
To find the time it takes for the compass to hit the ground, we can use the equation of motion for vertically falling objects:
h = 0.5 * g * t^2
Where:
h = height (vertical distance)
g = acceleration due to gravity (assumed to be 9.8 m/s^2 on Earth)
t = time
In this case, the initial height of the compass is 3.00 m above the ground, and it will fall downward with a speed of 2.50 m/s due to the upward motion of the balloon being canceled out. However, since the balloon is rising with a constant speed, the compass will not be affected by the horizontal velocity of the balloon.
Let's solve for t:
Let's assume that the downward direction is positive (since the compass is falling). We'll use the equation:
h = 0.5 * g * t^2
Rearranging the equation to solve for t:
t^2 = (2 * h) / g
t = sqrt((2 * h) / g)
Substituting the given values:
t = sqrt((2 * 3.00) / 9.8)
t = sqrt(0.6122)
t ā 0.782 s
So, it takes approximately 0.782 seconds for the compass to hit the ground after being dropped from the balloon.