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Vertically projected body-velocity after \(\frac{1}{n}^{th}\) of its time of ascent(\(t_{a}\))

we know that, \(\boxed{v=u+at}\)

here, v=v'

u=u

a=-g

t=\(\frac{t_{a}}{n}\)

when, t=\(t_{a}\)

v=o

a=-g

=>u-g\(t_{a}\)=0

=>u=g\(t_{a}\)__(1)

when, t=\(\frac{t_{a}}{n}\)

v'=u-g\(\frac{t{a}}{n}\)

v'=\(\frac{nu-gt_{a}}{n}\)

v'=\(\frac{nu-u}{n}\) [since, from (1)]

\(\boxed{v'=u\frac{n-1}{n}}\)

therefore, the velocity becomes \(\frac{n-1}{n}\) times its projected velocity

Note by Madhav Srirangan
1 year, 7 months ago

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You may \(\large \displaystyle \color{red}{\heartsuit}\) this! \(\ddot \smile\)

Rohit Udaiwal - 1 year, 7 months ago

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tq it'll be helpful, i hope.... :)

Madhav Srirangan - 1 year, 7 months ago

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