Question

A body is dropped from top if a tower. It covers 9/25 of total height of tower in last 1sec of its motion. Fond the height

Answer 1

Explanation:

Given

the height of tower = H

Total time taken to reach a ground = T

16H/25

0.5gT^2...... (1)

9H/25

0.5 ...... ( T-1) ^2 ......(2)

Divided above two equations

16/9= T/(T-1) ^ 2

4/3= T/(T-1)

3T= 4T -4

T=4 s

put the value of T= 4 in equation (1)

16H/25

0.5g×4^2

H= (25/16)0.5×9.8×16 (Cancel equation yourself better understand)

H= 122. 5 m

thank you

Answer 2

Given:

• The body covers $\dfrac{9}{25}$of distance in last 1sec

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Need to find:

• Height =?

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Solution:

Let the ball reach the ground in n seconds,

We know,

h$= ut + \dfrac{1}{2} g {t}^{2}$

Here u=0

Thus,

h= $\dfrac{1}{2} g {n}^{2}$

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We know,

$Snth = \dfrac{g}{2} (2n - 1)$

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Also given Snth= $\dfrac{9}{25} h$

$\implies \dfrac{g}{2}(2n - 1) = \dfrac{9}{25}h$

$\implies \dfrac{g}{2} (2n - 1) = \dfrac{9}{25} \times \dfrac{1}{2}g {n}^{2}$

$\implies 2n - 1 = \dfrac{9}{25} {n}^{2}$

$\implies 50n - 25 = 9 {n}^{2}$

$\implies 9 {n}^{2} - 50n + 25 = 0$

$\implies 9 {n}^{2} - 45n - 5n + 25 = 0$

$\implies 9n \: (n - 5) - 5(n - 5) = 0$

$\purple{\bold{\boxed {\implies n= 5 \: or \: n = \dfrac{5}{9}}}}$

$\rightarrow$ n can't be equal to 5/9 as it's less than 1 sec and is not considerable as the body takes 1sec to cover the 9/25th of height, thus it must take more than 1sec to complete it's journey.

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Considering g= 10 m/ ${sec}^{2}$

Height = $\dfrac{1}{2} \times 10 \times 5 \times 5 m$

$\red{\bold{\large{\boxed{\implies Height= 125m}}}}$

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