An Alaskan rescue plane traveling 42 m/s drops a package of emergency rations from a height of 117 m to a stranded party of explorers.
The acceleration of gravity is 9.8 m/s2 .
Where does the package strike the ground relative to the point directly below where it was released?
Answer in units of m
006 (part 2 of 3) 10.0 points
What is the horizontal component of the ve- locity just before it hits?
Answer in units of m/s
007 (part 3 of 3) 10.0 points
What is the vertical component of the velocity just before it hits? (Choose upward as the positive vertical direction)
Answer in units of m/s
Why did the package go to a party for explorers? Because it wanted to have a good time, of course! Now let's calculate the answers to your questions:
1. To find where the package strikes the ground relative to the point directly below where it was released, we can use the equation h = 1/2 * g * t^2, where h is the height, g is the acceleration due to gravity, and t is the time it takes to fall. We know that the height is 117 m and the acceleration due to gravity is 9.8 m/s^2. We can rearrange the equation to solve for time: t = sqrt(2h/g). Plugging in the values, we get t = sqrt(2 * 117 m / 9.8 m/s^2) ≈ 5.40 s.
Since the plane is traveling horizontally at a constant velocity, the package will also move horizontally with the plane. Therefore, the package will strike the ground at the same horizontal distance it was dropped from the point directly below where it was released. So the package strikes the ground relative to that point at 0 meters.
2. Because we know the horizontal component of the velocity remains constant, it will be 42 m/s.
3. The vertical component of the velocity just before it hits can be calculated using the equation v = g * t, where v is the vertical velocity and t is the time of fall. Plugging in the values, we get v = 9.8 m/s^2 * 5.40 s ≈ 52.92 m/s.
So, the package's vertical component of velocity just before it hits is approximately 52.92 m/s upwards.
To answer these questions, we need to use the equations of motion. Let's break the problem down step-by-step.
Step 1: Find the time it takes for the package to hit the ground.
We can use the equation:
h = 0.5 * g * t^2
Where:
h = height of the package (117 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the equation, we get:
t^2 = (2h) / g
Substituting the given values:
t^2 = (2 * 117 m) / 9.8 m/s^2
Calculating:
t^2 ≈ 23.8776
t ≈ √23.8776
t ≈ 4.8861 s
So, it takes approximately 4.8861 seconds for the package to hit the ground.
Step 2: Find the horizontal component of the velocity just before it hits.
The horizontal component of velocity remains constant throughout the motion because there is no horizontal acceleration.
Given that the plane is traveling at a constant velocity of 42 m/s, the horizontal component of the velocity just before the package hits the ground will also be 42 m/s.
Step 3: Find the vertical component of the velocity just before it hits.
We can use the equation:
v = u + gt
Where:
v = final velocity (unknown)
u = initial vertical velocity (0 m/s because the package was dropped)
g = acceleration due to gravity (9.8 m/s^2)
t = time (4.8861 s)
Substituting the values:
v = 0 + (9.8 m/s^2 * 4.8861 s)
v ≈ 47.876 m/s
So, the vertical component of velocity just before it hits the ground is approximately 47.876 m/s.
To find where the package strikes the ground relative to the point directly below where it was released, we can use the kinematic equation:
h = (1/2) * g * t^2
Where h is the height, g is the acceleration due to gravity, and t is the time.
First, let's find the time it takes for the package to hit the ground. We can use the formula:
h = (1/2) * g * t^2
Rearranging for t:
t = sqrt((2h) / g)
Substituting the given values:
t = sqrt((2 * 117 m) / (9.8 m/s^2))
t ≈ 5.53 seconds
Now, let's find the horizontal component of the velocity just before it hits the ground. Since there is no horizontal acceleration, the horizontal component of velocity remains constant throughout the motion. The horizontal velocity can be found using the formula:
v_x = v_initial * cos(theta)
Where v_initial is the initial velocity, and theta is the angle between the velocity and the horizontal direction. In this case, the angle is 0 degrees, so cos(theta) = 1.
Given that the Alaskan rescue plane is traveling at 42 m/s, the horizontal component of the velocity just before it hits the ground is:
v_x = 42 m/s
Finally, let's find the vertical component of the velocity just before it hits the ground. The vertical velocity can be found using the formula:
v_y = v_initial * sin(theta) - g * t
Where v_initial is the initial velocity, sin(theta) is the sine of the angle between the velocity and the horizontal direction, g is the acceleration due to gravity, and t is the time.
In this case, the initial velocity is 0 m/s (since the package was dropped), the angle is 90 degrees (straight down), and the time is 5.53 seconds.
v_y = 0 * sin(90 degrees) - (9.8 m/s^2) * 5.53 s
v_y ≈ -54.19 m/s
Therefore, the vertical component of the velocity just before it hits the ground is approximately -54.19 m/s. Note that the negative sign indicates the direction is downward.
time in air: t=sqrt(2gh)
horizontal distance: 42*time
horizonal velocity: same as plane
vertical final: g*timeinair