A small plane flying horizontally 150m above the ground at a constant speed of 120km/h is to drop a container with supplies to a point on the ground. How far from the point directly above the target should the supplies be dropped without giving them an initial push?

Question

A small plane flying horizontally 150m above the ground at a constant speed of 120km/h is to drop a container with supplies to a point on the ground. How far from the point directly above the target should the supplies be dropped without giving them an initial push?

Answer 1

To solve this problem, we can use the principle of projectile motion. The key concept to understand is that the horizontal and vertical motions are independent of each other.

In this case, the vertical motion involves the container being dropped from a height of 150 meters. The only force acting on the container vertically is gravity, which causes it to accelerate downward with a constant acceleration of approximately 9.8 m/s².

The horizontal motion is a constant speed of 120 km/h (33.33 m/s), meaning there is no horizontal acceleration.

To determine the horizontal distance from the point directly above the target where the supplies should be dropped, we need to figure out the time it takes for the container to fall vertically from a height of 150 meters.

We can use the equation for vertical displacement in projectile motion:

h = ut + (1/2)gt²,

where h is the height (150 meters), u is the initial vertical velocity (0 m/s because it's dropped), g is the acceleration due to gravity (-9.8 m/s² downwards), and t is the time taken.

Plugging in the values, we get:

150 = 0 + (1/2)(-9.8)t²

Simplifying the equation, we get:

-4.9t² = -150

Dividing both sides by -4.9:

t² = 30.61

Taking the square root of both sides:

t ≈ 5.53 seconds

So it takes approximately 5.53 seconds for the supplies to fall vertically from a height of 150 meters.

Now, to find the horizontal distance traveled by the plane in that time, we use the equation:

d = vt,

where d is the horizontal distance, v is the horizontal velocity (33.33 m/s), and t is the time (5.53 seconds).

Plugging in the values, we get:

d = (33.33)(5.53) ≈ 184.98 meters

Therefore, the supplies should be dropped approximately 184.98 meters horizontally from the point directly above the target without giving them an initial push.