An airplane traveling 1001 m above the ocean at 125km/h is going to drop a box of supplies to shipwrecked victims below. What is the horizontal distance between the plane and the victims when the box is dropped?

Woops!

125km/hr is 34.7m/s
so 34.7*14.3 is 496.21m

The time T that it will take the box to fall 1001 m, neglectring air resistance, is given by the equation

(g/2)T^2 = 1001
T^2 = 2002/9.8 = 204.3 s^2
T = 14.3 s
The distance that the box moves forward while it falls is V*T.

You will need to convert the 125 km/h speed to ___ m/s to get the distance in meters.

7.8m/s

Well, let me put on my comedic hat for this one. To calculate the horizontal distance between the plane and the victims, we need some information. Specifically, we need to know the angle at which the box is dropped and the time it takes for the box to reach the victims. Unfortunately, you didn't provide those details. So, we'll have to resort to some clownish speculation!

In a typical clown world, where flying boxes are as common as flying elephants, let's say the box is dropped straight down with a perfect 90-degree angle. In that case, the horizontal distance between the plane and the victims will be exactly zero, because gravity does its thing and brings the box straight down into the arms of our lucky shipwrecked pals.

However, in the non-clown universe we live in, we need more information to calculate the actual horizontal distance. So, do you have any additional details to share? Or should we just enjoy this imaginary sight of a flying box? 🤡

To find the horizontal distance between the plane and the victims when the box is dropped, we can use the concept of time and distance traveled.

First, we need to determine the time it takes for the box to reach the victims. To do this, we need to calculate the time it takes for the box to fall from the plane to the ocean.

We can use the equation of motion: s = ut + 1/2at^2. In this case, the initial velocity (u) is 0 m/s (since the box is dropped), the acceleration (a) is the acceleration due to gravity, which is approximately 9.8 m/s^2, and the distance traveled (s) is 1001 m.

Using this equation, we can rearrange it to find the time (t):
s = ut + 1/2at^2
1001 = 0*t + 1/2*9.8*t^2
1001 = 4.9*t^2
t^2 = 1001/4.9
t^2 ≈ 204.49
t ≈ √204.49
t ≈ 14.3 seconds

Now that we have the time it takes for the box to reach the ocean, we can find the horizontal distance traveled by the plane during this time.

The horizontal distance traveled is the product of the horizontal velocity and the time. The velocity of the plane is given as 125 km/h, but we need to convert it to m/s.

125 km/h = 125 * (1000 m/1 km) / (3600 s/1 h) = 34.72 m/s

Therefore, the horizontal distance traveled by the plane is:
distance = velocity * time
distance = 34.72 m/s * 14.3 s
distance ≈ 496.70 meters

So, the horizontal distance between the plane and the victims when the box is dropped is approximately 496.70 meters.

The time is 14.3s, btw..

Givens:

a=-9.81m/s^2
d=-1001m
vi=0m/s

d=vit+1/2at^2

-1001m=1/2-9.81m/s^2t^2
-1001m=-4.905m/s^2t^2
divide -1001m by -4.905m/s^2

204s^2=t^2
square root both sides

14.3s=t

125km/hr is 56m/s
56*14.3=
800.8m, that's the distance from the package dropped.