Question

A particle is dropped from the tower.In the last one second of motion, the particle determines the height of the tower as 9/25 then find the height of the tower?

Answer 1

$\huge{\sf{\boxed{\boxed{ANSWER}}}}$

Explanation:

THE HEIGHT OF TOWER CAN BE DETERMINED BY FOLLOWING STEPS:

Let the height of tower be 'h' .

During the last 1 sec it covers 9h/25 height.

we have the equation :

S=ut +1/2at²  (since the body is falling from a height a=g,u=0)

9h/25 = 0 +1/2*10*1²

9h=5*25

h=125/9

h=13.89 m

Answer 2

AnswEr :

⋆ In last second(say t second) it travel 9/25 of the tower height(say h metre) i.e. 9h/25 metre

⋆ So body travel (h - 9h/25) = 16h/25 metre in (t -1) second and h metre in t second.

• Now with the Formula :

$\rightarrow\mathsf{h = \dfrac{1}{2} \: g {t}^{2} }$⠀⠀ —(¡)

$\rightarrow\mathsf{\dfrac{16h}{25} = \dfrac{1}{2} \: g {(t - 1)}^{2} }$⠀⠀ —(¡¡)

• Putting the Value of h in (¡¡)

$\rightarrow\mathsf{\dfrac{16}{25} \times \cancel{\dfrac{1}{2}g} \: {t}^{2} = \cancel{\dfrac{1}{2} g} \: {(t - 1)}^{2} }$

$\rightarrow\mathsf{ \dfrac{16 {t}^{2} }{25} = {(t - 1)}^{2} }$

$\rightarrow\mathsf{16 {t}^{2} = 25( {t}^{2} + 1 - 2t) }$

$\rightarrow\mathsf{16 {t}^{2} = 25 {t}^{2} + 25 - 50t }$

$\rightarrow\mathsf{9 {t}^{2} - 50t + 25 = 0 }$

$\rightarrow\mathsf{9 {t}^{2} - 45t - 5t + 25 = 0 }$

$\rightarrow\mathsf{9t(t - 5) - 5(t - 5) = 0}$

$\rightarrow\mathsf{(9t - 5)(t - 5) = 0}$

$\rightarrow\mathsf{t = \dfrac{5}{9} \: \: or \: \: t = 5 }$

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• Particle Total Distance Travelled in 5 sec.

$\longrightarrow\mathsf{h = \dfrac{1}{2} \: g {t}^{2} }$

$\longrightarrow\mathsf{h = \dfrac{1}{ \cancel2} \times \cancel{10} \times {(5)}^{2} }$

$\longrightarrow\mathsf{h = 5 \times 25}$

$\longrightarrow\large\mathsf{h =125 \: m}$

⠀

$\large\therefore$ Height of the Tower is 125 m.