An airplane dropping supplies to northern villages that are isolated by severe blizzards and cannot be reached by land vehicles. The airplane is flying at an altitude of 785m and at a constant horizontal velocity of 53.5 m/s. At what horizontal distance before the drop point should the co-pilot drop the supplies so that they will land at the drop point?

(neglect air friction).

time to ground is 785/4.9=12.66

drop point 53.5*12.66 = 676.7 m before target

How long does it take to drop 785m?

4.9t^2 = 785

Then the horizontal distance traveled in that time is 53.5t

Well, if we neglect air friction, it sounds like we're also neglecting any potential distraction caused by clown pilots. Nevertheless, let me calculate the answer without clowning around too much.

We can use the horizontal velocity and time to determine the horizontal distance. Since the plane is flying at a constant horizontal velocity of 53.5 m/s, we know that the supplies will also be moving at this velocity.

To find the time it takes for the supplies to reach the ground, we need to consider the vertical motion. The supplies will fall from an altitude of 785m, so we can use the equation for free fall distance, h = (1/2)gt^2, where g is the acceleration due to gravity (9.8 m/s^2).

Rearranging the equation, we find that t^2 = 2h / g. Substituting the values, we get t^2 = 2 * 785 / 9.8, which simplifies to t^2 = 160.204.

Taking the square root of both sides, we find that t ≈ 12.663 seconds.

Since we know the horizontal velocity (53.5 m/s) and the time (12.663 seconds), we can now find the horizontal distance traveled. Multiplying these two values together, we get approximately 678.986 meters.

Therefore, to ensure that the supplies land at the drop point, the co-pilot should drop them approximately 678.986 meters before reaching the drop point.

Just be careful not to drop any clown noses or rubber chickens along the way. They might create a few laughs, but they won't be much use to those who need supplies!

To find the horizontal distance before the drop point where the supplies should be dropped, we need to first determine the time it takes for the supplies to reach the ground.

We can use the equation:

y = y0 + v0y * t + (1/2) * a * t^2

where:
- y is the vertical position (altitude) of the supplies (785m)
- y0 is the initial vertical position of the supplies (0m)
- v0y is the initial vertical velocity of the supplies, which is 0 m/s (since the supplies are dropped, not thrown)
- a is the acceleration due to gravity, which is approximately 9.8 m/s^2
- t is the time it takes for the supplies to reach the ground (unknown)

Plug in the values:

785 = 0 + 0 * t + (1/2) * 9.8 * t^2

Simplifying the equation:

785 = (1/2) * 9.8 * t^2

Multiply both sides by 2:

1570 = 9.8 * t^2

Divide both sides by 9.8:

t^2 = 1570 / 9.8

t^2 ≈ 160.2

Taking the square root of both sides:

t ≈ √160.2

t ≈ 12.65 seconds

Now that we know the time it takes for the supplies to reach the ground, we can find the horizontal distance traveled by the supplies using the equation:

d = v0x * t

where:
- d is the horizontal distance traveled by the supplies (unknown)
- v0x is the initial horizontal velocity of the supplies, which is 53.5 m/s (given)
- t is the time it takes for the supplies to reach the ground (12.65 seconds)

Plug in the values:

d = 53.5 * 12.65

Simplifying the equation:

d ≈ 677 meters

Therefore, the co-pilot should drop the supplies approximately 677 meters before the intended drop point.

To determine the horizontal distance at which the co-pilot should drop the supplies, we can use the equation of motion in the horizontal direction.

The equation for horizontal motion is:
distance = velocity × time

In this case, the velocity is the constant horizontal velocity of the airplane, 53.5 m/s. We want to find the time it takes for the supplies to reach the ground, so we need to determine the time it takes for the supplies to fall from the airplane to the ground.

To find this time, we can use the equation of motion in the vertical direction. Since initial vertical velocity is 0 m/s and the only force acting on the supplies is gravity, we can use the following equation:

vertical distance = (1/2) × acceleration due to gravity × time^2

The vertical distance is given by the altitude of the airplane, which is 785 m. The acceleration due to gravity is 9.8 m/s^2. Rearranging this equation, we can solve for time:

time = square root of ((2 × vertical distance) / acceleration due to gravity)

Substituting the given values:
time = square root of ((2 × 785 m) / 9.8 m/s^2)

Once we have the time it takes for the supplies to fall, we can use this time to calculate the horizontal distance using the equation of motion in the horizontal direction:

distance = velocity × time

Substituting the given values:
distance = 53.5 m/s × time

Now, let's calculate the values:

time = square root of ((2 × 785 m) / 9.8 m/s^2)
time = square root of (1570 m / 9.8 m/s^2)
time = square root of 160.204 m/s^2
time = 12.66 s (rounded to two decimal places)

distance = 53.5 m/s × 12.66 s
distance = 678.11 m (rounded to two decimal places)

Therefore, the co-pilot should drop the supplies approximately 678.11 meters before the drop point for them to land at the desired location.