Motion Under Gravity

EQUATIONS OF MOTION FOR FREELY FALLING BODY

Motion of all the dropped bodies falling rds the Earth when air resistance is ignored is mown as free fall.
For a freely falling body, u=0, a=+g
(i)v=u+at v=gt
(ii)\( S = ut + \frac{1} {2}at^2 \) \( \Rightarrow S = \frac{1} {2}gt^2 \)
(iii) \( v^2 - u^2 = 2aS \Rightarrow v^2 = 2gS \)
(iv) \( S_n = u + a\left[ {n - \frac{1} {2}} \right] \Rightarrow S_n = g\left[ {n - \frac{1} {2}} \right] \)

1.For a freely falling body, the ratio of distances travelled in 1 second, 2 seconds,   3 seconds,... =1:4:9:16.so on
2.For a freely falling body, the ratio of distances travelled in successive seconds 1:3:5:9 so on
3.The fraction of distance fallen in n^th second is \( \frac{{S_n }} {S} = \frac{{(2n - 1)}} {{n^2 }} \)
4.If s_n is the distance travelled by a body in n^th with uniform acceleration,and s_m is the distance travelled in m^th sec then
               \( S_n = u + \frac{a} {2}\left[ {2n - 1} \right],S_m = u + \frac{a} {2}\left[ {2m - 1} \right] \)  
                 \( \begin{gathered} S_n - S_m = a(n - m) \hfill \\ \therefore a = \frac{{S_n - S_m }} {{n - m}} \hfill \\ \end{gathered} \)    
In uniform acceleration motion, the distance travelled in every second increases by an amount equal to the magnitude of acceleration.