A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge H = 231 m below. If the plane is traveling horizontally with a speed of 227 km/h (63.1 m/s), how far in advance of the recipients (horizontal distance) must the goods be dropped?

433m

Suppose, instead, that the plane releases the supplies a horizontal distance of x = 325 m in advance of the mountain climbers. What vertical velocity (use the positive direction as upwards) should the supplies be given so that they arrive precisely at the climbers' position? And at what speed do the supplies land ?

To determine how far in advance the goods must be dropped, we can use the formula for horizontal distance:

𝑑 = 𝑣 × 𝑡

Where:
𝑑 = distance
𝑣 = horizontal velocity (given as 63.1 m/s)
𝑡 = time

First, we need to determine the time it will take for the supplies to reach the ground, considering only the vertical motion. We can use the equation for vertical displacement:

ℎ = 0.5 × 𝑔 × 𝑡²

Where:
ℎ = vertical displacement (given as 231 m)
𝑔 = acceleration due to gravity (approximately 9.8 m/s²)
𝑡 = time

Rearranging the equation, we get:

𝑡² = (2 × ℎ) ÷ 𝑔

Substituting the given values, we can find 𝑡:

𝑡² = (2 × 231 m) ÷ 9.8 m/s²
𝑡² = 47.04
𝑡 ≈ 6.86 s

Now, we can substitute the value of 𝑡 into the equation for horizontal distance:

𝑑 = 63.1 m/s × 6.86 s
𝑑 ≈ 432.97 m

Therefore, the goods must be dropped approximately 433 m in advance of the recipients.

For the second part of the question, we want to calculate the vertical velocity of the supplies so that they arrive precisely at the climbers' position.

Vertical velocity can be found using the equation:

𝑣_𝑖 = √(2 × 𝑔 × ℎ)

Where:
𝑣_𝑖 = initial vertical velocity
ℎ = vertical displacement (given as 231 m)
𝑔 = acceleration due to gravity (approximately 9.8 m/s²)

Substituting the values, we can calculate 𝑣_𝑖:

𝑣_𝑖 = √(2 × 9.8 m/s² × 231 m)
𝑣_𝑖 ≈ 66.93 m/s

Therefore, the supplies should be given an initial vertical velocity of approximately 66.93 m/s to arrive precisely at the climbers' position.

To calculate the speed at which the supplies land, we can use the equation:

𝑣 = √(𝑣_𝑖² + 2 × 𝑔 × ℎ)

Substituting the values, we can calculate 𝑣:

𝑣 = √(66.93 m/s)² + 2 × 9.8 m/s² × 231 m)
𝑣 ≈ 68.77 m/s

Therefore, the supplies will land at a speed of approximately 68.77 m/s.

To determine the vertical velocity and speed of the supplies when they arrive precisely at the climbers' position, we can use the following steps:

1. Calculate the time it takes for the supplies to fall vertically from the plane to the climbers' position:
- Use the equation: Δy = v₀yt + (1/2)gt², where Δy is the vertical distance, v₀y is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s²).
- Δy is 231 m (the height difference) and v₀y is unknown.
- Since the climbers are on the ground, Δy = 0, and the equation becomes: 0 = v₀yt + (1/2)gt².

2. Calculate the time it takes for the supplies to travel horizontally from the plane to the climbers' position:
- Use the formula: Δx = v₀xt, where Δx is the horizontal distance, v₀x is the initial horizontal velocity, and t is the time.
- Δx is 325 m (the horizontal distance) and v₀x is 63.1 m/s (the horizontal speed of the plane).

3. Equate the time from step 1 to the time from step 2 to find the vertical velocity:
- Set v₀yt + (1/2)gt² = v₀xt.
- Substitute the known values: (1/2)(-9.8)t² = 63.1t.
- Simplify the equation: -4.9t² - 63.1t = 0.
- Factor out t: t(-4.9t - 63.1) = 0.
- Solve for t: t = 0 (discard) or t = -63.1 / -4.9.

4. Calculate the vertical velocity:
- Substitute the value of t into v₀yt + (1/2)gt² = 0.
- v₀y(-63.1 / -4.9) + (1/2)(-9.8)(-63.1 / -4.9)² = 0.
- Simplify the equation: -v₀y(63.1 / 4.9) - 63.1(63.1 / 4.9) = 0.
- Solve for v₀y: v₀y = -63.1(63.1 / 4.9) / (63.1 / 4.9).

5. Calculate the speed at which the supplies land:
- Use the Pythagorean theorem: speed = √(v₀x² + v₀y²).
- Substitute the values of v₀x and v₀y into the equation to calculate the speed.

By following these steps, you should be able to determine the vertical velocity and speed at which the supplies should be given for them to arrive precisely at the climbers' position.

To fall 231 m, the time required is

t = sqrt(2H/g) = 6.87 seconds.
The supplies will travel forward
6.87 * 63.1 = 433.2 m
Your answer is correct.

For the second part, the time to fall must be changed to (325 m)/V = 5.15 seconds

The supplies must then be given an initial (negative) velocity v downwards so that it reaches the ground in less time. Solve this equation
-231 = -4.9 t^2 +v*t
= -130.0 + 5.15 v
v = -101/5.15 = -19.7 m/s

The supplies land at velocity v - g*t
where t = 5.15 s

Aerodynamic friction has been neglected.

It's going to hit the ground very fast.