Question

A stone is thrown vertically upward with a velocity of 40m/s and ia caught back taking 10m/s .calculate the maximum height reached total distances coverd by th stone... please answer with full solution. :-) (»:<br />

Answer 1

See the above attachment for your answer :-)

Hope this helps.....

Answer 2

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
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 .