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?
The question does not include part (a). I hope this is intended.
The problem can be summarized as follows:
Horizontal distance, H=425 m
Vertical distance, V=-235 m
horizontal velocity, h = 77.78 m/s
time to reach target, t = H/h = 5.464 s
The vertical component of velocity, v, is such that when the projectile is ejected upwards with a velocity v, it should reach V=-235 in 5.464 seconds.
The kinematics equation gives us:
V = vt + (1/2)(-g)t²
Solving for v give v=-16.2 m/s
The negative sign means the supplies should be given a kick downwards.
Check my calculations.
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To answer part (b) of the question, we need to calculate the time it takes for the supplies to reach the mountain climbers. We can use the horizontal distance and the horizontal speed of the plane to do this.
The horizontal distance traveled by the supplies is 425 m. The horizontal speed of the plane is 77.8 m/s.
The time it takes for the supplies to reach the climbers can be calculated using the formula: time = distance / speed.
Substituting the values, we have:
time = 425 m / 77.8 m/s
Simplifying, we find the time it takes for the supplies to reach the climbers is approximately 5.463 seconds.
Now, to answer part (c) of the question, we need to calculate the vertical velocity of the supplies so that they arrive precisely at the climbers' position.
The vertical distance between the plane and the climbers is 235 m.
We can use the formula for vertical distance traveled (d) in free fall:
d = (vo * t) + (0.5 * g * t^2)
where vo is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.
Since the supplies start from rest vertically (as they were dropped from the plane), we can simplify the formula to:
d = 0 + (0.5 * g * t^2)
We can rearrange the formula to solve for the initial vertical velocity (vo):
vo = (d - 0.5 * g * t^2) / t
Substituting the values, we have:
vo = (235 m - 0.5 * 9.8 m/s^2 * (5.463 s)^2) / 5.463 s
Simplifying, we find the vertical velocity of the supplies should be approximately 17.11 m/s downwards.
Finally, to answer part (d) of the question, we need to calculate the speed at which the supplies land.
The speed at landing will be the resultant of the horizontal speed of the plane and the vertical velocity of the supplies.
Using the Pythagorean theorem, we can find the speed at landing (v):
v = sqrt((horizontal speed)^2 + (vertical velocity)^2)
Substituting the values, we have:
v = sqrt((77.8 m/s)^2 + (17.11 m/s)^2)
Simplifying, we find the speed at which the supplies land is approximately 80.35 m/s.
To find the vertical velocity and speed of the supplies, we can split the problem into horizontal and vertical components.
(a) First, let's determine the time it takes for the supplies to reach the ground. We can use the vertical motion equation:
h = ut + (1/2)gt^2
Here, h is the height (235 m) and g is the acceleration due to gravity (-9.8 m/s^2). Since the initial vertical velocity (u) is zero (the supplies are released horizontally), the equation simplifies to:
235 = (1/2)(-9.8)t^2
Solving for t:
t^2 = (2 * 235) / 9.8
t^2 = 48.0
t = sqrt(48.0)
t ≈ 6.93 s
(b) Now, let's determine the needed vertical velocity for the supplies to land precisely at the climbers' position. We can use the horizontal motion equation:
d = vt
Here, d is the horizontal distance (425 m) and v is the horizontal velocity of the plane (77.8 m/s). Rearranging the equation to solve for the time taken (t):
t = d / v
t = 425 / 77.8
t ≈ 5.46 s
Since the supplies need to hit the climbers and the climbers take 6.93 s to reach the ground, the vertical velocity of the supplies needs to be such that they take the remaining time (6.93 - 5.46 = 1.47 s) to drop 235 m.
Using the equation 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 taken, we rearrange to solve for vf:
vf = vi + gt
0 = vi + (-9.8)(1.47)
vi = 9.8 * 1.47
vi ≈ 14.4 m/s
So, the supplies need to be given a vertical velocity of approximately 14.4 m/s upwards.
(c) Finally, to find the speed at which the supplies land, we can use the Pythagorean theorem:
speed = sqrt((horizontal speed)^2 + (vertical speed)^2)
Given that the horizontal speed is 77.8 m/s and the vertical speed is 14.4 m/s:
speed = sqrt((77.8)^2 + (14.4)^2)
speed = sqrt(6048.84 + 207.36)
speed = sqrt(6256.2)
speed ≈ 79.1 m/s
Therefore, the supplies will land with a speed of approximately 79.1 m/s.