Question

An astronaut on the moon drops a feather from rest (v = 0). The acceleration due to gravity is 1.67 m/s​ 2​ . If the feather begins 2 meters above the moon’s surface, what will be its final position after falling for 1.5 seconds?

Answer 1

Given :

Initial velocity , u = 0 m/s .

Acceleration due to gravity on moon , $g_m=1.67\ m/s^2$ .

Height , h = 2 m .

To Find :

Final position after falling for 1.5 seconds .

Solution :

We know , by equation of motion :

$s=ut+\dfrac{at^2}{2}$

Here , $a = g_m$ .

So , equation will transform by :

$s=ut+\dfrac{g_mt^2}{2}\\\\s=0+\dfrac{1.67\times 1.5^2}{2}\ m\\\\s=1.88\ m$

Therefore , the height form moon's surface is 1.88 m .

Hence , this is the required solution .