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?
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?
To determine the horizontal distance the goods must be dropped in advance of the recipients, we can use the formula for horizontal distance traveled by an object in motion:
distance = velocity * time
In this case, the velocity of the plane is given as 63.1 m/s. To find the time it takes for the goods to reach the recipients, we need to calculate the time of flight. The plane is traveling horizontally, so its vertical velocity is 0 m/s.
Using the formula for time of flight in free fall:
time = sqrt(2 * height / gravity)
where height is the vertical distance of 231 m and gravity is approximately 9.8 m/s^2, we can substitute the values to find the time of flight.
time = sqrt(2 * 231 / 9.8) = 10.756 seconds
Now we can find the horizontal distance traveled by the goods:
distance = velocity * time = 63.1 m/s * 10.756 s = 678.9416 m
Therefore, the goods must be dropped approximately 679 meters in advance of the recipients.
To determine the vertical velocity that the supplies should be given so that they arrive precisely at the climbers' position, we can use the formula for vertical displacement in free fall:
displacement = initial velocity * time + (1/2) * acceleration * time^2
Since the vertical displacement should be -231 m (negative because it is downward), the initial velocity is unknown, the acceleration is -9.8 m/s^2, and the time of flight is 10.756 seconds, we can rearrange the formula to solve for the initial velocity:
-231 = initial velocity * 10.756 + 0.5 * -9.8 * (10.756)^2
Simplifying and solving for the initial velocity:
-231 = 10.756 * initial velocity - 54.78 * 10.756
10.756 * initial velocity = 54.78 * 10.756 - 231
initial velocity = (54.78 * 10.756 - 231) / 10.756
initial velocity = 549.726 - 21.317 = 528.4096 m/s
Therefore, the supplies should be given an initial vertical velocity of approximately 528.41 m/s (upwards) to arrive precisely at the climbers' position when released 325 m in advance.
To find the distance in advance that the goods must be dropped, we can use the formula:
distance = speed * time
In this case, the speed of the plane is 63.1 m/s and the vertical distance (H) is 231 m. We need to find the time it takes for the supplies to reach the climbers.
Using the equation of motion for vertical motion:
H = (1/2) * g * t^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time.
Rearranging the equation, we have:
t = sqrt(2H / g)
Substituting the given values:
t = sqrt(2 * 231 / 9.8)
t ≈ 7.56 s
Now, we can find the horizontal distance by multiplying the time by the horizontal velocity of the plane:
distance = speed * time
distance = 63.1 m/s * 7.56 s
distance ≈ 477.94 m
Therefore, the goods must be dropped approximately 477.94 meters in advance of the recipients.
Next, to find the vertical velocity required for the supplies to arrive precisely at the climbers' position, we can use the formula:
Vf = Vi + gt
where Vf is the final vertical velocity, Vi is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.
Since the supplies need to be given an upward velocity to counteract the effect of gravity, the initial vertical velocity (Vi) would be zero (assuming the supplies are released from rest).
Therefore, we have:
Vf = 0 + 9.8 m/s^2 * 7.56 s
Vf ≈ 74.33 m/s (upwards)
So, the supplies should be given a vertical velocity of approximately 74.33 m/s (upwards) to arrive precisely at the climbers' position.