A hot-air balloon is descending at a rate of 1.7 m/s when a passenger drops a camera. If the camera is 44 m above the ground when it is dropped, how long does it take for the camera to reach the ground?
h=v₀t+gt²/2
gt²+2 v₀t-2h=0
10 t²+3.4t-90 =0
t= {- 3.4±sqrt(11.56+3600)}/20=
= { - 3.4 ± 60.1}/20
t = 2.83 s
To find the time it takes for the camera to reach the ground, we can use the equation of motion for free fall. The equation is given by:
h = 1/2 * g * t^2,
where:
h is the height,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time it takes to fall.
Since the hot-air balloon is descending at a rate of 1.7 m/s, the effective speed of the camera will be the difference between the descending speed and the camera's own falling velocity. So, the effective speed of the camera will be:
1.7 m/s + 0 m/s (since the camera is dropped) = 1.7 m/s.
Now, let's plug the known values into the equation and solve for t:
44 m = 1/2 * 9.8 m/s^2 * t^2.
Rearranging the equation:
t^2 = (2 * 44 m) / 9.8 m/s^2,
t^2 = 88 m / 9.8 m/s^2,
t^2 = 8.97.
Taking the square root of both sides:
t ≈ √8.97,
t ≈ 2.99.
Therefore, it takes approximately 2.99 seconds for the camera to reach the ground.