Question

A stone is dropped from the adge of a roof.find the speed of the stone at 2 second after the stone was dropped.

Answer 1

finding acceleration
$a = vf - vi \div t$
vf =0m/s
vi=0m/s
t=2sec

hence
a=0.5m/s^2

now, for distance
$2as = vf {}^{2} - vi {}^{2}$
2(0.5)s=0^2-0^2
1s=0
s=0/1
s = 0m
for speed
$s = distance \div time$

speed=0/2
hence s=0m/s

Answer 2

$\huge\tt{\underline{\underline{Given:-}}}$

• A stone is dropped from the edge of roof

$\huge\tt{\underline{\underline{To\:find:-}}}$

• Speed of the stone after 2 seconds

$\huge\tt{\underline{\underline{Solution:-}}}$

For stone ,

u = 0 m/s

t = 2 sec

a = g = 9.8 m/s²

V = ? , S = ?

From Second equation of motion ,

$\large\rm\bold{\underline{\boxed{S\:=\:ut+\frac{1}{2}\times\:at^2}}}$

$\large\rm{S\rightarrow{Distnace\:(or)\:Displacement\:travelled}}$

$\large\rm{u\rightarrow{Initial\:velocity}}$

$\large\rm{t\rightarrow{Time}}$

$\large\rm{\implies{S\:=\:0(2)+\frac{1}{2}\times\:(9.8)(2^2)}}$

$\large\rm{\implies{S\:=\:0+4.9(4)}}$

$\large\rm{\implies{S\:=\:19.6\:m}}$

∴ Distance travelled by stone after 2 sec is 19.6 m

Now ,

From Third equation of motion ,

$\large\rm\bold{\underline{\boxed{v^2-u^2\:=\:2as}}}$

$\large\rm{S\rightarrow{Distnace\:(or)\:Displacement\:travelled}}$

$\large\rm{u\rightarrow{Initial\:velocity}}$

$\large\rm{v\rightarrow{Final\:velocity}}$

By substituting the values ,

$\large\rm{v^2-(0)^2\:=\:2(9.8)(19.6)}$

$\large\rm{\implies{v^2\:=\:384.16}}$

$\large\rm{\implies{v\:=\:\sqrt{384.16}}}$

$\large\rm{\implies{v\:=\:19.6\:m/s}}$

∴ The velocity of stone after 2 sec is equal to 19.6 m/s

$\huge\tt{\underline{\underline{Note:-}}}$

★ Equations of motion are valid only when accleration of particle is constant

$\huge\tt{\underline{\underline{\green{Additional\:Information:-}}}}$

❃ First equation of motion is given by ,

$\large\rm\bold{\underline{\boxed{V\:=\:u+at}}}$

$\large\rm{V\rightarrow{Final\:velocity}}$

$\large\rm{u\rightarrow{Initial\:velocity}}$

$\large\rm{t\rightarrow{Time}}$

$\large\rm{a\rightarrow{acceleration}}$

❃ Fourth equation of motion is given by ,

$\large\rm\bold{\underline{\boxed{S_n\:=\:u+a(\:n-\frac{1}{2})}}}$

$\large\rm{S\rightarrow{Distnace\:(or)\:Displacement\:in\:nth\:sec}}$

$\large\rm{u\rightarrow{Initial\:velocity}}$

$\large\rm{a\rightarrow{acceleration\:of\:particle}}$

$\large\rm{n\rightarrow{The\:particular\:second\:in\:a\:journey}}$