A helicopter traveling upward at 120 m/s drops a package from a height of 500 meters. To the nearest second how long does it take to hit the ground?
I believe that there is mistake in your given data: the velocity has to be 12 m/s
There are two motions of the package:
1. upward motion with the initial velocity of 12 m/s
h=vₒ²/2g =144 /2•9.8 =7.34 m
H=500+7.34 =507.34 m
2. Dowmward motion (free fall)
t=sqrt(2H/g)=sqrt(2•507.34/9.8) =10.18 s.
If the speed is 120 m/s, calculate using your numbers.
To find the time it takes for the package to hit the ground, we can use the equation of motion. The distance the package falls is 500 meters, and the initial velocity is given as 0 m/s (since the package is dropped). We need to find the time, so we can rearrange the equation to solve for time:
S = ut + (1/2)at^2
Where:
S = distance
u = initial velocity
a = acceleration
t = time
Since the package is dropped, the initial velocity is 0 m/s. The acceleration due to gravity is approximately 9.8 m/s². Plugging in the values:
500 = 0*t + (1/2)*9.8*t^2
Simplifying the equation:
500 = 4.9*t^2
Dividing both sides of the equation by 4.9:
t^2 = 102.04
Taking the square root of both sides:
t ≈ √102.04
t ≈ 10.1 seconds
So, to the nearest second, it takes approximately 10 seconds for the package to hit the ground.