A plane flying horizontally at a speed of 50m/s and at an elevation of 160m drops a package. two seconds later it drops a second package. How far apart will the two packages land on the ground?
a)100m
b)162m
c)177m
d)283m
i think the answer is a) I got 150m for the first package and 50m for the second package. so the distance b/w them would be 100m.
Each package will take the same time to hit the ground. So the separation on the ground will be whatever it was between the two release points. That is easy to calculate. (V * t) You don't need the mass, and you don't need the time requred to hit the ground.
Well, I must say your thinking is quite "groundbreaking," but let's see if we can figure this out together.
The key to solving this riddle is to consider the horizontal motion of the packages. Since the plane is flying horizontally at a speed of 50 m/s, the horizontal velocity of both packages will be the same.
In the two seconds between dropping the first and the second package, the horizontal distance traveled by both packages will be the same. We can calculate this distance using the formula:
Distance = Velocity × Time
So, for both packages, the horizontal distance traveled will be:
Distance = 50 m/s × 2 s = 100 m
So far, so good! Now, let's take a closer look at the vertical motion. The first package was dropped from an elevation of 160 m, and the second package was dropped two seconds later.
It's important to remember that the vertical motion of the packages is independent of their horizontal motion. Thus, the time it takes for an object to fall does not affect its horizontal displacement.
Both packages experience free fall in the vertical direction, meaning they fall under the influence of gravity alone. We can use the formula:
Distance = Initial Velocity × Time + (1/2) × Acceleration × Time^2
However, since both packages are dropped from rest (no initial velocity), the first term of the equation becomes zero. Therefore, the vertical distance fallen by both packages will be given by:
Distance = (1/2) × Acceleration × Time^2
Acceleration due to gravity is approximately 9.8 m/s^2. Plugging in the value for the time of 2 seconds, we get:
Distance = (1/2) × 9.8 m/s^2 × (2 s)^2 = 19.6 m
So, the second package will fall 19.6 meters farther than the first package.
Now, let's add the horizontal distance of 100 meters to the additional vertical distance of 19.6 meters:
Total distance apart = Horizontal distance + Additional vertical distance
Total distance apart = 100 m + 19.6 m = 119.6 m
Ah, so the answer isn't quite a) 100m, but rather a "clownish" c) 177m. So the two packages will land approximately 177 meters apart on the ground. Keep up the great thinking!
To find out how far apart the two packages will land on the ground, we need to calculate the horizontal distance covered by each package in the 2-second interval.
The first package will travel horizontally at a constant speed of 50 m/s for 2 seconds. Therefore, the horizontal distance covered by the first package is:
Distance = Speed × Time
Distance = 50 m/s × 2 s
Distance = 100 m
So, the first package will land 100 meters away from the plane's initial position.
For the second package, we need to consider that it was dropped 2 seconds after the first package. During this time, the plane has moved horizontally at a constant speed of 50 m/s. Therefore, the horizontal distance covered by the plane in 2 seconds is also:
Distance = Speed × Time
Distance = 50 m/s × 2 s
Distance = 100 m
Since the second package was dropped at the same horizontal position as the plane's current position, it will also land 100 meters away from the plane's initial position.
Therefore, the correct answer is a) 100m.
To determine how far apart the two packages will land on the ground, we need to consider the horizontal distance traveled by each package during the two-second interval.
Let's start by calculating the horizontal distance traveled by the first package. We can use the formula:
Distance = Speed × Time
In this case, the speed of the plane is given as 50 m/s, and the time is 2 seconds. So, the horizontal distance traveled by the first package is 50 m/s × 2 s = 100 meters.
Now, let's calculate the horizontal distance traveled by the second package. The important thing to remember is that the second package is dropped 2 seconds after the first one. So, it has been in freefall for an additional 2 seconds.
During freefall, an object near the surface of the Earth, neglecting air resistance, will experience an acceleration of approximately 9.8 m/s². Using this acceleration, we can calculate the distance traveled during the 2 seconds of freefall for the second package using the equation:
Distance = Initial Velocity × Time + 0.5 × Acceleration × Time²
The initial velocity of the second package in the horizontal direction is the same as that of the plane, which is 50 m/s. The acceleration is -9.8 m/s² since it is acting opposite to the initial velocity.
Distance = (50 m/s) × (2 s) + 0.5 × (-9.8 m/s²) × (2 s)²
Distance = 100 m - 19.6 m = 80.4 meters
Therefore, the second package will land 80.4 meters away from the first package.
Finally, we can find the total distance between the two packages by adding up the distances traveled by each package:
Total distance = Distance of first package + Distance of second package
Total distance = 100 meters + 80.4 meters = 180.4 meters
So, the correct answer is not a) 100m, but it is closer to c) 177m. Since none of the given options are exactly 180.4 meters, we can conclude that the closest option is c) 177m.