Also got lost on this question:
A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235m below. If the plane is traveling horizontally with a speed of 250 km/h (69.4 m/s), (a) how far in advance of the recipients (horizontal distance) must the goods be dropped. (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 package has the same horizontal velocity component as the plane. (Note- if you drop a bomb at low altitude, turn your airplane immediately after release)
Fall of 235 m
235 = (1/2)g t*2 = 4.9 t^2
t = 6.93 seconds to fall
horizontal problem
d = u t
d = 69.4 * 6.93 = 481 meters
==========================
if at 425 meters then t = 425m/69.4 m/s
= 6.12 seconds instead of 6.93
we better give it a negative Vi so it gets down faster
-235 = Vi (6.12) - 4.9 (6.12^2)
-235 = 6.12 Vi -184
-51.2 = 6.12 Vi
Vi = - 8.37
To solve this question, we will use the principles of projectile motion. Let's break down each part of the question and explain how to find the answers.
(a) To determine how far in advance of the recipients the goods must be dropped, we need to find the time it takes for the supplies to reach the climbers' location.
First, we need to find the time it takes for the supplies to fall vertically down 235m. We can use the formula:
s = ut + (1/2)gt^2
where s is the distance (235m), u is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s^2).
Since the supplies are dropped vertically, the initial vertical velocity (u) is 0. Therefore, the equation simplifies to:
s = (1/2)gt^2
235 = (1/2)(-9.8)t^2
Solving for t:
t^2 = 235 / (0.5 * 9.8)
t^2 = 47.959
t = √47.959
t ≈ 6.925 seconds.
Now, we need to find the horizontal distance covered by the plane during this time. We can use the formula:
distance = speed * time
distance = 69.4 m/s * 6.925 s
distance ≈ 480.0375 m
Therefore, the goods must be dropped approximately 480 meters in advance of the recipients.
(b) To determine the vertical velocity of the supplies so that they arrive precisely at the climbers' position, we need to consider the horizontal distance covered by the plane.
First, calculate the time it takes for the supplies to travel a horizontal distance of 425m:
time = distance / speed
time = 425 m / 69.4 m/s
time ≈ 6.116 seconds.
Next, we need to calculate the initial vertical velocity required for the supplies to reach the climbers' position in this time. We can use the formula:
vertical velocity = (vertical displacement - (0.5 * g * t^2)) / t
Since the climbers are 235m below, the vertical displacement is -235m.
vertical velocity = (-235 - (0.5 * -9.8 * 6.116^2)) / 6.116
vertical velocity ≈ -19.022 m/s
Therefore, the supplies should be given a vertical velocity of approximately -19.022 m/s (downward) to arrive precisely at the climbers' position.
(c) To find the speed at which the supplies land in the latter case, we need to calculate the magnitude of the final velocity. We can use the formula:
final velocity = √(horizontal velocity^2 + vertical velocity^2)
Since the supplies are dropped vertically, the horizontal velocity remains constant at 69.4 m/s.
final velocity = √(69.4^2 + (-19.022)^2)
final velocity ≈ 72.465 m/s
Therefore, the speed at which the supplies land in the latter case is approximately 72.465 m/s.
To solve this question, we need to use the principles of projectile motion. We'll break it down step-by-step for parts (a), (b), and (c) of the question.
(a) To find the horizontal distance the goods should be dropped in advance, we need to determine the time it takes for the goods to fall from the plane to the mountain climbers.
1. First, we need to find the time of flight (t) of the goods. We can use the equation:
t = vertical distance / vertical component of velocity
Since the vertical distance is 235m and the vertical component of velocity is zero (because the plane is traveling horizontally), the time of flight is:
t = 235m / 0 m/s = undefined
The time of flight is undefined because the goods will never reach the mountain climbers vertically.
2. However, we can calculate the horizontal distance (d) the plane travels during this flight time using the horizontal component of velocity:
d = horizontal component of velocity x time
d = 69.4 m/s x t
Since we found that t is undefined, it means that the goods need to be dropped right above the mountain climbers. Therefore, the horizontal distance is zero.
So, the goods need to be dropped directly above the recipients, with a horizontal distance of zero.
(b) In this case, the supplies are released 425m in advance of the mountain climbers. We need to find the vertical velocity of the supplies to ensure they reach the climbers' position.
1. The time of flight (t) can be found using the equation:
t = horizontal distance / horizontal component of velocity
Given that the horizontal distance is 425m, and the horizontal component of velocity is 69.4 m/s, we have:
t = 425m / 69.4 m/s = 6.12s (rounded to two decimal places)
2. We can then find the required vertical velocity by rearranging the equation for vertical distance:
vertical distance = vertical component of velocity x time + (0.5 x gravity x time^2)
Since vertical distance is -235m (negative because it is downward), and the time is 6.12s, and gravity is 9.8 m/s^2, we can solve for the vertical component of velocity:
-235m = vertical component of velocity x 6.12s - (0.5 x 9.8 m/s^2 x (6.12s)^2)
After substituting values and simplifying, we can solve for the vertical component of velocity.
(c) To find the speed at which the supplies land in the second case, we need to calculate the resultant velocity (magnitude of velocity).
1. The resultant velocity can be found using the equation:
resultant velocity = √(horizontal component of velocity)^2 + (vertical component of velocity)^2
Substitute the values of the horizontal component of velocity (69.4 m/s) and the vertical component of velocity, found in part (b), into the equation and solve for the resultant velocity.
After following these steps, you should be able to find the answers for parts (a), (b), and (c) of the question.