A rescue plane wants to drop supplies to isolated mountains climbers on Rocky ridge 235m below. If the plane is traveling horizontally with a speed of 250 km/h (69.4m/s), how far in advance of the recipients (horizontal distance) must the good be dropped?
"Dropped" (Vyo=0)
To determine the horizontal distance in advance of the recipients where the supplies must be dropped, we can use the kinematic equation for horizontal motion:
d = v * t
Where:
- d is the horizontal distance,
- v is the horizontal velocity of the plane, and
- t is the time taken for the supplies to reach the recipients.
Since the plane is traveling horizontally with a speed of 69.4 m/s, the horizontal velocity (v) is 69.4 m/s.
For the supplies to reach the mountain climbers 235 m below, we need to calculate the time taken (t). Since the supplies are being dropped vertically, the vertical motion can be analyzed separately.
Using the equation for vertical motion:
d = v * t + (1/2) * g * t^2
Where:
- d is the vertical distance (235 m below),
- v is the initial vertical velocity (0 m/s, dropped),
- g is the acceleration due to gravity (9.8 m/s^2), and
- t is the time taken.
Since the initial vertical velocity (Vyo) is 0, the equation simplifies to:
d = (1/2) * g * t^2
Rearranging the equation to solve for time (t):
t = √(2d / g)
Substituting the values of d = 235 m and g = 9.8 m/s^2:
t = √(2 * 235 / 9.8)
t ≈ √(47.96)
t ≈ 6.93 s
Now that we have the time taken for the supplies to reach the recipients, we can calculate the horizontal distance (d) using the previously mentioned equation:
d = v * t
Substituting the values of v = 69.4 m/s and t ≈ 6.93 s:
d = 69.4 * 6.93
d ≈ 481.042 m
Therefore, the supplies must be dropped approximately 481.042 meters in advance of the recipients horizontally.
To calculate the horizontal distance the supplies must be dropped in advance, we need to consider the time it takes for the supplies to reach the ground.
Since the plane is traveling horizontally, the vertical speed of the supplies when dropped will be governed by the force of gravity. The vertical velocity of the supplies at impact will be the terminal velocity, which can be calculated using the formula:
Vf^2 = Vi^2 + 2gh
where Vf is the final velocity (terminal velocity), Vi is the initial velocity (which is zero), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height (235 m).
Using these values, we can solve for Vf:
Vf^2 = 0^2 + 2 * 9.8 * 235
Vf^2 = 2 * 9.8 * 235
Vf^2 = 4598
Vf ≈ 67.8 m/s
Now that we know the vertical velocity of the supplies when dropped, we can calculate the time it takes for the supplies to reach the ground using the equation:
h = (1/2) * g * t^2
where h is the height (235 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
Substituting the known values:
235 = (1/2) * 9.8 * t^2
470 = 9.8 * t^2
t^2 = 470 / 9.8
t^2 ≈ 47.96
t ≈ √47.96
t ≈ 6.93 s
Now we can calculate the horizontal distance the supplies must be dropped in advance. The distance traveled horizontally can be found using the equation:
d = v * t
where d is the distance, v is the horizontal velocity (69.4 m/s), and t is the time (6.93 s).
Substituting the known values:
d = 69.4 * 6.93
d ≈ 480.54 m
Therefore, the goods must be dropped approximately 480.54 meters in advance of the recipients horizontally.