Question

A particle is dropped from a tower. it is found that it Travels 45 m in the last second of its journey. then the height of the Tower is

Answer 1

S=u+1/2g(2t-1)
45=0+1/2*10(2*t-1)
then s= 5 sec.
now
h=1/2gt^2
h=1/2*10*5^2=125m.


hopw this will be help u
if u like please mark as brainlist

Answer 2

Answer:

• $\sf{h = 125\:m}$

Explanation:

${\underline{\underline{\sf{\red{{\bigstar}\:\:\:Given:-}}}}}$

$\tiny\:\:\:\:\bullet\:\:\:\sf\green{Particle\:travels\: 55m\: in\: the\: last \:second\: of \:its\: journey }$

${\underline{\underline{\sf{\orange{{\bigstar}\:\:\:ToFind:-}}}}}$

$\tiny\:\:\:\:\bullet\:\:\:\sf\purple{Height\: of \: the \:tower}$

${\underline{\underline{\sf{\blue{{\bigstar}\:\:\: Solution:-}}}}}$

$\tiny\dag\:\sf{Let \ the \ total\: time \ of \ journey \ be \ n \ seconds}$

$\star\:{\underline{\sf{\red{Distance\: travelled \ in \ n^{th} \ sec}}}}:-$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ S_{n} = u + \frac{a}{2}(2n-1) }$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ 45 = 0+\frac{10}{2}(2n-1)}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{n = 5\:sec}$

$\star\:{\underline{\sf{\purple{Height\: of \: the \: tower}}}}:-$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{h = \frac{1}gt^2}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{h = \frac{1}{2} \times 10\times 5^2}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf{ h = 125\:m}$