A stone is dropped from the top of a tower of height ‘H’ ,in last second of journey it travels a distance of  0.36 H. Then   height H and time of fall of stone  is g=10m/s2

Let ‘n’ is last sec Sn is distance covered in last secwe know, S=Ut+12at2, here  displacement=s=H;  initial velocity=U=0;  acceleration due to gravity=a=g; total time elapsed=t=n; substitute given values in the equation      H=12gn2---(1)  Distance travelled before the last second  =H−0.36H=0.64Hdistance covered before last second=12gn−12=0.64H substitute equation (1) ⇒12gn−12=0.6412gn2 ⇒0.64n2=n−12 ⇒0.64n2=n−12⇒64100n2=n−12⇒8n=10n−10⇒2n=10⇒n=5⇒n=5=total time of flight The time of fall is 5sec  h=12gn2 ⇒h=121052 ⇒h=2502 ⇒h=125m=height of tower