A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. The plane is traveling horizontally with a speed of 280 km/h (77.8 m/s).
(b) Suppose, instead, that the plane releases the supplies a horizontal distance of 425 m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climbers' position?
(c) With what speed do the supplies land in the latter case?
To find the vertical velocity of the supplies and their speed of landing, we can use the equations of motion for projectile motion.
Let's start with part (b) of the question.
(b) To find the vertical velocity the supplies should be given, we need to determine the time it takes for the supplies to reach the climbers' position. Since the plane is traveling horizontally, the time it takes for the supplies to reach the climbers will be the same as the time it takes for the supplies to travel horizontally.
The horizontal distance traveled by the supplies is 425 m, and the horizontal speed of the plane is given as 77.8 m/s. Therefore, we can use the formula:
time = distance / speed
In this case, the distance is 425 m and the speed is 77.8 m/s. Plugging in the values, we get:
time = 425 m / 77.8 m/s
time ≈ 5.464 s
Now that we have the time, we can use it to find the vertical velocity of the supplies. The vertical distance the supplies need to cover is 235 m, and the time taken is 5.464 s. We can use the formula:
vertical velocity = vertical distance / time
Plugging in the values, we get:
vertical velocity = 235 m / 5.464 s
vertical velocity ≈ 43.02 m/s
Therefore, the supplies should be given a vertical velocity of approximately 43.02 m/s in the upwards direction so that they arrive precisely at the climbers' position.
(c) Now, let's find the speed at which the supplies land when released 425 m in advance of the mountain climbers. Since we know the horizontal and vertical velocities separately, we can use the Pythagorean theorem to find the resulting speed of the supplies.
The horizontal velocity of the supplies is the same as the plane's speed, which is given as 77.8 m/s. The vertical velocity is 43.02 m/s (upwards).
Using the Pythagorean theorem, we can calculate the resulting speed:
speed = √(horizontal velocity^2 + vertical velocity^2)
speed = √(77.8^2 + 43.02^2)
speed ≈ √(6047.24 + 1852.0404)
speed ≈ √7899.2804
speed ≈ 88.87 m/s
Therefore, the supplies will land with a speed of approximately 88.87 m/s when released 425 m in advance of the mountain climbers.