Question

An object is dropped from a helicopter flying horizontally at 360 km/h. It falls from a height of 2 km and reaches the ground in 20 seconds. What is the displacement of the package relative to the helicopter's position when it was dropped?

• A. 2 km
• B. \( 2\sqrt{2} \) km
• C. 4 km
• D. \( 8 \) km

Hint:

Remember, the displacement relative to the helicopter includes both the vertical fall and horizontal motion of the object.

Answer 1

The Correct Option is B

The object is dropped horizontally from the helicopter. In this case, the horizontal velocity of the helicopter will be the horizontal velocity of the object upon release. The vertical velocity can be found using the equation for free fall. The horizontal displacement is given by: \[ d = v_x t \] Where \( v_x \) is the horizontal velocity and \( t \) is the time. From the problem, we know the horizontal velocity \( v_x = 360 \, \text{km/h} \), and time \( t = 20 \, \text{s} \). Converting \( v_x \) to m/s: \[ v_x = 360 \times \frac{1000}{3600} = 100 \, \text{m/s} \] Thus, the horizontal displacement is: \[ d = 100 \times 20 = 2000 \, \text{m} = 2 \, \text{km} \] So the displacement relative to the helicopter is \( 2\sqrt{2} \, \text{km} \).