Question

A cargo helicopter, descending steadily at a speed of 2.3 m/s, releases a small package. Let upward be the positive direction for this problem.
(a) If the package is 71 m above the ground when it is dropped, how long does it take for the package to reach the ground?
1 s

(b) What is its velocity just before it lands? (Indicate the direction with the sign of your answer.)
2 m/s

Answer 1

Explanation:

Given that,

Given that,

Speed of the helicopter, $u_x=2.3\ m/s$

Vertical velocity of the helicopter, $u_y=0$

(a) Height of the package, h = 71 m

We need to find the time taken for the package to reach the ground. It can be calculated using second equation of motion as :

$h=u_yt+\dfrac{1}{2}at^2$

Here a = g

$h=\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2h}{g}}$

$t=\sqrt{\dfrac{2\times 71}{9.8}}$

t = 3.80 seconds

(b) We can find the vertical velocity of package. It is given by :

$v_y=u_y-gt$

$v_y=-gt$

$v_y=-9.8\times 3.8=-37.24\ m/s$

Let v is the velocity of the package just before it lands. It is given by :

$v=\sqrt{v_x^2+v_y^2}$

$v=\sqrt{2.3^2+37.24^2}$

v = 37.31 m/s

Hence, this is the required solution.