A hot-air balloon is rising upward with a constant speed of 2.95 m/s. When the balloon is 3.30 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?
Consider the compass: you know the initial vertical velocity, its initial height, and gravitational acceleration.
hfinal=hinitial+ vi*time -4.9 t^2
solve for t. hfinal is 0, hinitial is 3.30
5m/s
To find the time it takes for the compass to hit the ground, we can use the equation of motion for vertical motion:
š = š£šššš” + 0.5šš”Ā²
where:
š = distance travelled (in this case, 3.30 m)
š£š = initial velocity (in this case, 0 m/s since the compass is dropped)
š = acceleration (in this case, the acceleration due to gravity, which is approximately 9.8 m/sĀ²)
š” = time (what we're trying to find)
Rearranging the equation, we get:
0.5šš”Ā² + š£šššš” - š = 0
Now we can substitute the known values into the equation:
0.5(9.8 š”)Ā² + (0)(š”) - 3.30 = 0
Simplifying the equation, we have:
4.9š”Ā² - 3.30 = 0
Now, we can solve this quadratic equation using the quadratic formula:
š” = (-š Ā± ā(šĀ² - 4šš)) / (2š)
In this case, š = 4.9, š = 0, and š = -3.30. Plugging these values into the formula, we get:
š” = Ā± ā((0)Ā² - 4(4.9)(-3.30)) / (2(4.9))
Simplifying further, we have:
š” = Ā± ā(0 + 64.68) / 9.8
š” = Ā± ā(64.68) / 9.8
Since time cannot be negative, we take the positive square root:
š” = ā(64.68) / 9.8
Evaluating this using a calculator, we find:
š” ā 1.20 s
Therefore, it takes approximately 1.20 seconds for the compass to hit the ground.