Question

A box containing food supplies for a refugee camp was dropped from a helicopter flying horizontally at a constant elevation of \(500 . \mathrm{m}\). If the box hit the ground at a distance of \(150 . \mathrm{m}\) horizontally from the point of its release, what was the speed of the helicopter? With what speed did the box hit the ground?

Short answer

Answer: The speed of the helicopter is 21.0 m/s, and the speed at which the box hits the ground is 72.1 m/s.

Step-by-step solution

Finding the time of free fall

Using the following kinematic equation for the vertical motion, we can find the time it takes for the box to fall 500 m: \(h = \frac{1}{2}gt^2\) where \(h=\) Elevation \(g=\) Acceleration due to gravity \((9.81 m/s^2)\) \(t=\) Time of free fall We can now solve for \(t\): \(500 = \frac{1}{2}(9.81)t^2\) \(t^2 = \frac{500\cdot 2}{9.81}\) \(t = \sqrt{\frac{1000}{9.81}}\) We can now calculate the time using the given values: \(t = \sqrt{\frac{1000}{9.81}} \approx 7.146 \ \text{s}\)

Calculating the speed of the helicopter

Now that we have the time it takes for the box to hit the ground, we can find the horizontal speed of the helicopter using the following equation: \(speed = \frac{distance}{time}\) where \(distance=\) Horizontal distance traveled by the box \((150m)\) The speed of the helicopter can be calculated as: \(speed = \frac{150}{7.146} \approx 21.0 \ \text{m/s}\)

Calculating the final vertical velocity of the box

Now we need to find the vertical velocity of the box just before hitting the ground. We can use the following kinematic equation: \(v_f = gt\) where \(v_f =\) Final vertical velocity The vertical velocity just before the box hits the ground can be calculated as: \(v_f = (9.81)(7.146) \approx 70.0 \ \text{m/s}\)

Calculating the overall speed of the box

Lastly, we will use the Pythagorean theorem to find the overall speed of the box as it hits the ground, combining its horizontal speed (which is equal to the helicopter speed) and its final vertical speed. \(overall\_speed = \sqrt{horizontal\_speed^2 + vertical\_speed^2}\) Plugging in the speeds, we get: \(overall\_speed = \sqrt{(21)^2 + (70)^2} \approx 72.1 \ \text{m/s}\) In summary, the speed of the helicopter was \(21.0 \ \text{m/s}\), and the speed of the box hitting the ground was \(72.1 \ \text{m/s}\).

Key concepts

Kinematic Equations

Kinematic equations are at the heart of describing the motion of an object in physics. They help us understand how an object moves under the influence of forces, particularly gravity, without the need to consider mass and friction. In projectile motion problems like the one given, these equations let us divide the motion into horizontal and vertical components.

For vertical motion, when an object is in free fall, the equation used is \(h = \frac{1}{2}gt^2\), where \(h\) is the distance fallen, \(g\) is the acceleration due to gravity, and \(t\) is the time. By rearranging this equation, one can solve for the time it takes for an object to fall a certain distance, as in the box example.

Free Fall

Free fall is a particular case of motion where gravity is the only force acting on an object. All objects in free fall (in the absence of air resistance) accelerate downwards at the same rate, regardless of their mass. This constant acceleration is denoted by \(g\) and typically has a value of approximately \(9.81 m/s^2\) on Earth.

When the box was dropped from the helicopter, it was in free fall, and we calculated how long it took to reach the ground using the kinematic equation. The vertical velocity of the box just before impact was found using \(v_f = gt\), another kinematic equation specific to objects in free fall.

Pythagorean Theorem in Physics

The Pythagorean theorem is a principle from geometry that also finds great use in physics, particularly in problems involving projectile motion. It provides a method to calculate the resultant magnitude of a vector when its perpendicular components are known.

In our example, the box's final speed as it hits the ground can be found by combining its horizontal speed (which is the same as the helicopter's speed) and its final vertical speed using the Pythagorean theorem: \(overall\_speed = \sqrt{horizontal\_speed^2 + vertical\_speed^2}\). This delivers a more complete picture of the motion, as we consider both the horizontal and vertical aspects of the box's journey to the ground.