A hot-air balloon is rising upward with a constant speed of 4.00 m/s. When the balloon is 4.77 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?
u=4m/s
when constant speed, acceleration a=0
distance s=4.77
time t=?
s=u*t+1/2*a*t^2
s=u*t because a=0
t=s/u=4.77/4
t=1.77 sec.
To determine how much time elapses before the compass hits the ground, we can use the equation of motion for vertical motion with constant acceleration. Since we are dealing with a uniform upward velocity, the acceleration is zero. The equation of motion in this case is:
h = ut + (1/2)gt^2
Where:
h = height (4.77 m)
u = initial velocity (0 m/s, as the compass is dropped)
g = acceleration due to gravity (-9.8 m/s^2, taking the direction into account)
t = time
Given that the initial velocity (u) is zero, the equation simplifies to:
h = (1/2)gt^2
Rearranging the equation to solve for time (t):
t^2 = (2h) / g
t = sqrt((2h) / g)
Substituting the given values into the equation:
t = sqrt((2 * 4.77) / 9.8)
Simplifying the expression:
t = sqrt(0.4887)
Calculating the square root:
t ā 0.7 seconds
Therefore, the compass will hit the ground approximately 0.7 seconds after it is dropped from the balloon.