Question

A particle is dropped from the top of tower of height 4.9 m . The velocit of particle at ground is ?

Answer 1

according to newton's Third kinematics eqn
square of velocity = square ( u) + 2( as)
= 0 + 2*9.8*4.9
= 96.04 m/s
hence its velocity must be 96.04m/s
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Answer 2

GIVEN :-

• A particle is dropped from height 4.9 m

TO FIND :-

• The velocity of particle at ground

SOLUTION :-

u = 0 (Since the particle is dropped)

h = 4.9 m

a = g = 9.8 m/s²

From third equation of motion ,

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

Where ,

• v is final velocity
• u is initial velocity
• a is acceleration of particle
• s is distance covered

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

$\large\rm{\rightarrow{v^2\:=\:(19.6)^2}}$

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

∴ The velocity of the particle at ground is 19.6 m/s

$\large\rm{\underline{\underline{\green{Additional\:Information:-}}}}$

❃ First equation of motion is given by ,

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

where ,

• V is final velocity
• u is initial velocity
• a is acceleration
• t is time

❃ Second equation of motion is given by ,

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

Where ,

• S is distance covered
• u is initial velocity
• t is time
• a is accleration of particle