Answer to Question #272058 in Mechanics | Relativity for secret

Question

A chopper drops boxes of relief goods to a group of people stranded in an island. The goods were released 20.0 m above the sea and landed 40.0 m from a point exactly below where the goods were released. What was the velocity of the goods when they were released? What was the velocity of the chopper?

Answer 1

$h = 20m \newline s = 40m\newline g = 9.8\frac{m}{s^2} \text{Before the released of the cargo: }\newline v_{chopper}= v_{cargo}\newline \text{After the release of the cargo:}\newline \vec v_{cargo}= \vec v_x+\vec v_y\newline \vec v_x = \vec v_{chopper}\newline v_y = v_{y_0}+gt\newline v_{y_0}= 0\newline t = \sqrt{\frac{2h}{g}}= \sqrt{\frac{2*20}{9.8}}=2.02s\newline v_y = gt = 9.8*2.02\approx 19.80\frac{m}{s}\newline s = v_xt\newline v_x=\frac{s}{t}= \frac{40}{2.02}\approx19.80\frac{m}{s}$

$v = \sqrt{v_x^2+v_y^2}=28\frac{m}{s}$

$\text{Answer: }$

$v_{chopper} = 19.80\frac{m}{s}$

$v'_{cargo} = 19.80\frac{m}{s}- \text{during separation from the helicopter}$

$v_{cargo} = 28.0\frac{m}{s}- \text{during the landing of the cargo}$