**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**

## Comments

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TopNewestYou may \(\large \displaystyle \color{red}{\heartsuit}\) this! \(\ddot \smile\) – Rohit Udaiwal · 7 months, 3 weeks ago

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– Madhav Srirangan · 7 months, 3 weeks ago

tq it'll be helpful, i hope.... :)Log in to reply