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

#Mechanics

Easy Math Editor

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Comments

You may $\large \displaystyle \color{#D61F06}{\heartsuit}$ this! $\ddot \smile$

Rohit Udaiwal - 5 years, 1 month ago

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

madhav srirangan - 5 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Vertically projected body-velocity after 1nth\frac{1}{n}^{th}n1​th of its time of ascent(tat_{a}ta​)

Comments

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