Answer to Question #168894 in Physics for Khalifa

Question

A helicopter is descending vertically at a constant speed of 3 ms^-1. A sandbag is released from the helicopter. The sandbag hits the ground 5.0 s later. What was the height of the helicopter above the ground at the time the sandbag was released?

Answer 1

Let's take the downwards as the positive direction. Then, we can find the height of the helicopter above the ground at the time the sandbag was released from the kinematic equation:

$$y=v_0t+\dfrac{1}{2}gt^2,$$ $$y=3\ \dfrac{m}{s}\cdot5\ s+\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(5\ s)^2=137.5\ m.$$