Question

a body is dropped freely under gravity from the top of a tower of height 8000 CM calculate the term to it to reach the ground ​

Answer 1

Answer:

As the body is dropped, its initial velocity is 0. Let the time taken to reach the ground be 't'. ∴ Time taken to reach the ground is 4.02 s

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Answer 2

Step-by-step explanation:

$\mathtt{ \implies \: s = ut + \frac{1}{2} at ^{2} }$

$\mathtt{let \: s = 8000. \: \: u = 0 .\: \: \: \: a = 9.8 \: \frac{m}{s} }$

$\mathtt{ \implies8000 = \: 0 \times t + \frac{1}{2} \times 9.8 \times {t}^{2} }$

$\mathtt{ \implies8000 = \frac{9.8}{2} \times {t}^{2} }$

$\mathtt{ \implies8000 = 4.9 \times {t}^{2} }$

$\mathtt{ \implies {t}^{2} = \frac{8000}{4.9} }$

$\mathtt{ \implies {t}^{2} = 1632.653061}$

$\mathtt{ \implies \:t=±40.406102</p><p></p><p></p><p> }$

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