A cargo plane is moving with a horizontal velocity of
vx = +223 m/s
at a height of
y = 980 m
above level ground as shown in the figure below when it releases a package. Ignoring air resistance, how much time will it take the package to reach the ground? (Express your answer to the nearest tenth of a second.)
Well, let's do some math and see if we can calculate that! We can determine the time it takes for the package to reach the ground by using the equation of motion in the vertical direction.
The equation is: y = y0 + v0yt + (1/2)gt²
Since the initial vertical velocity, v0y, is zero (since the package is released from rest), we can simplify the equation to: y = y0 + (1/2)gt²
Let's plug in the values:
y = 980 m (the initial height)
g = 9.8 m/s² (acceleration due to gravity)
Now, we just need to find t (the time it takes for the package to reach the ground). Rearranging the equation, we get:
t = √(2(y - y0) / g)
Plugging in the values:
t = √(2(0 - 980) / 9.8)
Crunching the numbers, we find that it takes approximately 14.3 seconds for the package to reach the ground. Rounded to the nearest tenth of a second, our answer is 14.3 seconds.
Now, that's a long time for a package to freefall! I hope there's nothing fragile in there!
To find the time it takes for the package to reach the ground, we can use the equation of motion for free fall:
y = y0 + v0y * t + 1/2 * a * t^2
Where:
y = final displacement (equals 0 for the package when it reaches the ground)
y0 = initial displacement (height of the cargo plane = 980 m)
v0y = initial vertical velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time
Plugging in the known values:
0 = 980 + 0 * t + 1/2 * (-9.8) * t^2
Simplifying the equation:
0 = 980 - 4.9 * t^2
Rearranging the equation:
4.9 * t^2 = 980
Dividing both sides by 4.9:
t^2 = 200
Taking the square root of both sides:
t = sqrt(200) ≈ 14.1 s
Therefore, it will take approximately 14.1 seconds for the package to reach the ground.
To find the time it takes for the package to reach the ground, we can use the equation of motion:
y = y0 + v0yt + (1/2)at^2
Where:
y is the final position (height of the ground),
y0 is the initial position (height of the cargo plane),
v0y is the initial vertical velocity (in this case, 0 since the cargo plane is moving horizontally),
a is the acceleration in the vertical direction (in this case, acceleration due to gravity, which is approximately 9.8 m/s^2),
t is the time.
First, let's find the initial vertical velocity v0y. Since the cargo plane is moving horizontally with a velocity of vx = +223 m/s, there is no vertical component to its velocity. Therefore,:
v0y = 0
Next, let's substitute the values into the equation:
0 = 980 + 0 + (1/2)(9.8)t^2
Simplifying the equation:
(1/2)(9.8)t^2 = -980
(9.8)t^2 = -1960
t^2 = -1960 / 9.8
t^2 = -200
t ≈ √(-200)
t ≈ 14.1 seconds
Note: The negative sign indicates that the package is moving in the opposite direction of the positive y-axis. However, we can ignore the negative sign since we are only interested in the magnitude of the time.
So, it will take approximately 14.1 seconds for the package to reach the ground.