A passenger in a helicopter traveling upwards at 27 m/s accidentally drops a package out the window. If it takes 17 seconds to reach the ground, how high to the nearest meter was the helicopter when the package was dropped?
d = Vo*t + 0.5*9.8t^2,
d = -27*17 + 0.5*9.8(17)^2,
= -459 + 1416 = 957 m.
The neg. initial velocity means the helicopter is moving in opposite direction of the free falling object.
To find out the height of the helicopter when the package was dropped, we can use the equation of motion. In this case, since the helicopter is traveling upwards, the initial velocity (u) would be positive.
Let's break down the information given:
- Initial velocity (u) = 27 m/s (upwards)
- Time taken (t) = 17 seconds
- Final velocity (v) = 0 m/s (at the topmost point of the trajectory)
- Acceleration (a) can be assumed to be the acceleration due to gravity, which is approximately 9.8 m/s².
The equation we can use here is:
v = u + at
Since the final velocity (v) is zero when the package reaches its maximum height, we can rearrange the equation to solve for the time taken to reach the highest point:
0 = 27 + (9.8)t
-27 = 9.8t
t ≈ -2.755 seconds
We can ignore this negative time value since it doesn't have any physical significance. Now that we have the time taken (t), we can calculate the displacement (s) using the formula:
s = ut + (1/2)at²
Plugging in the values:
s = (27 * 17) + (1/2)(9.8)(17)²
s ≈ 459 + (1/2)(9.8)(289)
s ≈ 459 + 1414.9
s ≈ 1873.9 meters
Therefore, the height to the nearest meter when the package was dropped from the helicopter is approximately 1874 meters.