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 solve part (b) of the problem, we need to determine the vertical velocity of the supplies in order for them to reach the mountain climbers.
We can begin by analyzing the horizontal motion of the plane. The supplies will take the same amount of time as the plane to travel the horizontal distance of 425 m. We can use the equation:
distance = speed × time
Rearranging the equation, we have:
time = distance / speed
Substituting the given values, we get:
time = 425 m / 77.8 m/s
≈ 5.46 s
Now, let's focus on the vertical motion of the supplies. We know that the supplies will be dropped 235 m below the plane. Therefore, the time the supplies take to fall this distance is the same as the time it takes for the plane to travel horizontally.
Using the equation for the distance traveled in free fall:
distance = (1/2) × acceleration × time^2
Since we are considering vertical motion, we can substitute the acceleration due to gravity, which is approximately 9.8 m/s^2. Rearranging the equation, we get:
time = √(2 × distance / acceleration)
= √(2 × 235 m / 9.8 m/s^2)
≈ 7.23 s
From our analysis, we can see that the supplies will take approximately 7.23 seconds to reach the mountain climbers. During this time, they must travel horizontally for 5.46 seconds.
Now, we can calculate the vertical velocity of the supplies. Since velocity is defined as the change in distance divided by the change in time, we have:
vertical velocity = distance / time
= (235 m) / (7.23 s)
≈ 32.48 m/s
Since the supplies are being dropped below the plane, the vertical velocity should be negative. Thus, the supplies should be given a vertical velocity of approximately -32.48 m/s (downward) to arrive precisely at the climbers' position.
Moving on to part (c) of the problem, we need to determine the speed at which the supplies will land in the latter case.
The speed at which the supplies land is the resultant of their horizontal and vertical velocities. We have already determined the vertical velocity as -32.48 m/s (downward).
To calculate the horizontal velocity, we can use the equation:
horizontal velocity = distance / time
= (425 m) / (5.46 s)
≈ 77.71 m/s
Using the Pythagorean theorem, the total speed of the supplies can be found:
speed = √(horizontal velocity^2 + vertical velocity^2)
= √((77.71 m/s)^2 + (-32.48 m/s)^2)
≈ 85.11 m/s
Therefore, the supplies will land with a speed of approximately 85.11 m/s in this case.