An airplane traveling 1007 m above the ocean at 150 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?

change velocity to m/s

how long does it take to fall 1007?
h=1/2 g t^2 solve for t.

then, what horizonal distance is covered in t seconds: distance=V*t calculate distance.

To solve this problem, we need to consider the motion of the airplane and the box separately.

(a) How many seconds before the plane is directly overhead should the box be dropped?
We can use the concept of time and displacement in order to determine when the box should be dropped.

The vertical displacement of the box, from the plane to the victims, is equal to the height of the plane above the ocean, which is 1007 meters.

The horizontal distance covered by the airplane when the box is dropped is equal to the speed of the plane multiplied by the time taken.

Given:
Vertical displacement (h) = 1007 meters
Speed of the plane (v) = 150 km/h

We need to convert the speed of the plane to meters per second:
1 km/h = (1000 m)/(3600 s)
So, 150 km/h = (150 * 1000 m)/(3600 s) = 41.67 m/s (rounding off to 2 decimal places)

Now we can calculate the time it takes for the plane to be directly overhead:
h = (1/2) * g * t^2, where g is the acceleration due to gravity (approx. 9.8 m/s^2) and t is the time taken.

Re-arranging the equation to solve for t:
t^2 = (2h)/g
t = sqrt((2h)/g)

Substituting the given values:
t = sqrt((2 * 1007 m) / (9.8 m/s^2))
t = sqrt(2014/9.8) s
t = sqrt(205.51) s
t ≈ 14.34 s (rounding off to 2 decimal places)

Therefore, the box should be dropped approximately 14.34 seconds before the plane is directly overhead.

(b) What is the horizontal distance between the plane and the victims when the box is dropped?
The horizontal distance covered by the airplane can be calculated using the formula: distance = speed × time.

Given:
Time taken (t) = 14.34 seconds
Speed of the plane (v) = 41.67 m/s (rounded off from km/h to m/s)

Substituting the values:
Distance = 41.67 m/s × 14.34 s
Distance ≈ 597.2 meters (rounding off to 1 decimal place)

Therefore, the horizontal distance between the plane and the victims when the box is dropped is approximately 597.2 meters.

To answer these questions, we can use the basic principles of kinematics and consider the motion of the airplane and the box separately.

(a) To determine how many seconds before the plane is directly overhead the box should be dropped, we need to calculate the time it takes for the box to reach the ground from the given height.

We can use the equation of motion:
\[s(t) = ut + \frac{1}{2}at^2\]
Where:
- \(s(t)\) is the displacement or height at time \(t\)
- \(u\) is the initial velocity
- \(a\) is the acceleration
- \(t\) is the time

Given:
- \(s(t) = 1007\) m (height above the ocean)
- \(u = 0\) m/s (since the box is dropped and has no initial velocity)
- \(a = g = 9.8\) m/s² (acceleration due to gravity)

Substituting the values:
\[1007 = 0t + \frac{1}{2}(9.8)t^2\]

Simplifying the equation:
\[4.9t^2 = 1007\]

Dividing both sides of the equation by 4.9:
\[t^2 = \frac{1007}{4.9}\]

Taking the square root of both sides:
\[t = \sqrt{\frac{1007}{4.9}}\]

Using a calculator, we find:
\[t \approx 10.05\] seconds

Therefore, the box should be dropped approximately 10.05 seconds before the plane is directly overhead.

(b) To find the horizontal distance between the plane and the victims when the box is dropped, we can use the equation of motion for horizontal motion:
\[s = ut\]
Where:
- \(s\) is the distance (horizontal)
- \(u\) is the velocity
- \(t\) is the time

Given:
- \(u = 150\) km/h = \(\frac{150 \times 1000}{3600}\) m/s (converting km/h to m/s)
- \(t = 10.05\) s (time calculated from part a)

Substituting the values:
\[s = (150 \times \frac{1000}{3600}) \times 10.05\]

Simplifying the equation:
\[s \approx 418.75\] meters

Therefore, the horizontal distance between the plane and the victims when the box is dropped is approximately 418.75 meters.