# 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

2 years, 2 months ago

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

- 2 years, 2 months ago

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

- 2 years, 2 months ago