Question

9. A stone is thrown vertically upwards with a velocity 40 m/s and is caught back. Taking g=10 m/s², calculate the maximum height reached by the stone. What is the net displacement and
the total distance covered by the stone?​

Answer 1

$\huge\bold{HELLO\:MATE}$

you did mistake in typing ,

" is caught back . taking g = 10m/s² "

A stone is thrown with speed 40 m/s , at highest point velocity of stone = 0 m/sec

use , kinematics equation ,

V² = U² + 2aS

where V = 0

U = 40 m/s

a = -g m/s² = -10 m/s²

0 = (40)² -2 × 10 × S

$\huge\boxed{S\: =\: 80 \:m }$

hence maximum hight is reached by stone = 80 m .

now, stone comes back to intial point.

hence

total distance covered by stone = 80 ( in upward motion ) + 80( in downward motion ) = 160 m .