A hot-air balloon is rising upward with a constant speed of 2.20 m/s. When the balloon is 2.50 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?
Vi = 2.2
initial height = 2.5
h = 2.5 + 2.2 t - 4.9 t^2 = 0 when hits ground
4.9 t^2 - 2.2 t - 2.5 = 0
t = [ 2.2 +/-sqrt (4.84 + 22) ] / 9.8
t = [ 2.2 + 5.18 ]/ 9.8 (ignore - time)
t = .753 seconds
To find the time it takes for the compass to hit the ground, we first need to determine the height from which it was dropped.
Since the hot-air balloon is rising with a constant speed, the height of the balloon above the ground increases over time. We know that the vertical velocity is 2.20 m/s, and we want to find the time it takes for the compass to hit the ground, starting from a height of 2.50 m.
Using the equation of motion for free fall:
h = h0 + V0t + (1/2)gt^2,
where
h is the final height,
h0 is the initial height,
V0 is the initial velocity (which is 0 since the compass was dropped),
t is the time,
g is the acceleration due to gravity (approximately -9.8 m/s^2 for downward motion).
Given that h is 0 (since it hits the ground) and h0 is 2.50 m, the equation becomes:
0 = 2.50 + 0t - (1/2)(9.8)t^2.
Simplifying the equation, we have:
(1/2)(9.8)t^2 = 2.50.
Now we can solve for t.
Multiplying both sides of the equation by 2:
9.8t^2 = 5.
Dividing both sides by 9.8:
t^2 = 5/9.8.
Taking the square root of both sides:
t = sqrt(5/9.8) ā 0.71 s.
Therefore, it takes approximately 0.71 seconds for the compass to hit the ground.