An airplane traveling 1004 m above the ocean at 130 km/h is going to drop a box of supplies to shipwrecked victims below.

(a) How many seconds before the plane is directly overhead should the box be dropped?

(b) What is the horizontal distance between the plane and the victims when the box is dropped?

Vo = 130km/h = 130,000m/3600s = 36.11m/sw.

a. h = 0.5g*t^2.
1004 = 4.9t^2, t = ?.

b. d = Vo*t = 36.11 * t,

To calculate the answers, we need to use the equations of motion for a falling object and the formula for calculating horizontal distance. Let's break down each question step by step:

(a) To find the time it takes for the box to reach the victims, we can use the equation of motion for a falling object:

s = ut + (1/2)gt^2

where:
s = vertical distance (1004 m in this case)
u = initial velocity (0 m/s as the box is dropped)
g = acceleration due to gravity (approx. 9.8 m/s^2)
t = time (unknown)

Rearranging the equation to solve for t:

t = sqrt((2s) / g)

Plugging in the values:

t = sqrt((2 * 1004) / 9.8) ≈ 14.21 seconds

Therefore, the box should be dropped 14.21 seconds before the plane is directly overhead.

(b) To find the horizontal distance between the plane and the victims, we can use the formula for horizontal distance:

d = vt

where:
d = horizontal distance (unknown)
v = horizontal velocity (130 km/h)
t = time (14.21 seconds)

First, we need to convert the given horizontal velocity from km/h to m/s. Since 1 km = 1000 m and 1 hour = 3600 seconds, the conversion factor is 1000/3600:

v = (130 km/h) * (1000 m/km) * (1 h/3600 s) ≈ 36.11 m/s

Plugging in the values:

d = (36.11 m/s) * (14.21 s) ≈ 513.88 meters

Therefore, the horizontal distance between the plane and the victims when the box is dropped is approximately 513.88 meters.