Question

A freely falling particle, falling from a tower of height h,
covers a distance h/2 in the last second of its motion. The
height of the tower is (Take g = 10 m/s2)
(a) 58 m (b) 50 m (c) 60 m (d) 55 m​

Answer 1

if we have g=10m/sec2

then ,

s=ut+1/2at2

1st sec body cover the distance/hight

s=(0)1+1/2(10)(1)2

s=5m

1+2nd sec coverd distance is

s=(0)2+1/2(10)(2)2

s=20m

then

distance coverd only 2nd sec =20-5

=15

and this ratio also mantain

like 1st sec cover distance =5

2nd sec ''''''''''''''''''''''''''''''''''''=15

3rd sec ''''''''''''''''''''''''''''''''''''''=25

if g=9.8m/sec then body cover the hight is 3rd sec =aproximate 25

so it seamed that the 50m of tower hight is right answer

ok sorry for any mistake

Answer 2

Answer:

heY... MaTe... ❤❤❤

AnsweR... 58 .. Yrr... ..

58 answeR... fuLL.. piC..Nahi.. aaYa..

soRrY...