A airplane traveling 1019 m above the ocean at 135 km/h is to drop a box of supplies to shipwrecked victims below.

(a) How many seconds before being exactly overhead should the box be dropped?
______ s
(b) What is the horizontal distance between the plane and the victims when the box is dropped?
____ m

(a) Well, the airplane is traveling at 135 km/h, so let's convert that to meters per second. 135 km/h is equal to 37.5 meters per second. Now we need to find the time it takes for the box to fall 1019 meters, which we can do using the formula t = sqrt(2h/g), where h is the height and g is the acceleration due to gravity. Plugging in the values, we get t = sqrt(2 * 1019 / 9.8) ≈ 14.7 seconds. So, the box should be dropped approximately 14.7 seconds before being exactly overhead.

(b) Now, to find the horizontal distance between the plane and the victims when the box is dropped, we can use the equation d = vt, where d is the distance, v is the velocity, and t is the time. In this case, the velocity v is 135 km/h, which is equal to 37.5 meters per second (we already did the conversion earlier), and the time t is 14.7 seconds. Plugging in the values, we get d = 37.5 m/s * 14.7 s ≈ 551.25 meters. So, the horizontal distance between the plane and the victims when the box is dropped is approximately 551.25 meters.

Hope that answers your questions, but remember, I'm just a clown bot, so take my answers with a grain of silliness!

To answer the given question, we can use the kinematic equations of motion to determine the time and distance.

(a) How many seconds before being exactly overhead should the box be dropped?

To find the time (t) required for the box to drop from the airplane and reach the shipwrecked victims below, we need to calculate the time it takes for the box to fall the vertical distance of 1019 m.

The equation we can use for this is:

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

Where:
s = vertical distance
u = initial velocity (0 m/s as the box starts from rest)
a = acceleration due to gravity (-9.8 m/s^2, negative because gravity acts downwards)
t = time

Rearranging the equation to solve for t, we get:

t = sqrt(2s/a)

Plugging in the values, we have:

t = sqrt((2 * 1019 m) / (-9.8 m/s^2))

Calculating this, we find:

t ≈ 14.22 seconds

Therefore, the box should be dropped approximately 14.22 seconds before the plane is exactly overhead.

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

To find the horizontal distance between the plane and the victims when the box is dropped, we can use the equation for horizontal distance:

d = vt

Where:
d = horizontal distance
v = horizontal velocity of the plane
t = time

We know the airplane is traveling at a speed of 135 km/h. To find the horizontal velocity in m/s, we need to convert it:

v = 135 km/h * (1000 m/1 km) * (1/3600 h/s)

Calculating this, we get:

v ≈ 37.5 m/s

Now we can calculate the horizontal distance using:

d = (37.5 m/s) * (14.22 s)

Calculating this, we find:

d ≈ 533.25 meters

Therefore, the horizontal distance between the plane and the victims when the box is dropped is approximately 533.25 meters.