Given :
Initial velocity , u = 0 m/s .
Acceleration due to gravity on moon , [tex]g_m=1.67\ m/s^2[/tex] .
Height , h = 2 m .
To Find :
Final position after falling for 1.5 seconds .
Solution :
We know , by equation of motion :
[tex]s=ut+\dfrac{at^2}{2}[/tex]
Here , [tex]a = g_m[/tex] .
So , equation will transform by :
[tex]s=ut+\dfrac{g_mt^2}{2}\\\\s=0+\dfrac{1.67\times 1.5^2}{2}\ m\\\\s=1.88\ m[/tex]
Therefore , the height form moon's surface is 1.88 m .
Hence , this is the required solution .
