Question

a body is projected vertically up and the total distance traveled by it is s. then total time of flight will be

Answer 1

When body is projected against gravity the initial speed of projection is let say "u" and deceleration due to gravity is "g"

now by the formula of kinematics

$v_f^2 - v_i^2 = 2 a s$

now we will find the distance traveled by it till it stops

so we will have

$0^2 - u^2 = 2(-g)s$

$s = \frac{u^2}{2g}$

now we can say that total distance of motion will be double of this because here the distance it will move up will be same as the distance it will come back to ground again.

$S = \frac{u^2}{g}$

from above equation we can find the speed "u"

$u = \sqrt{gS}$

now the total time of flight can be found similar way

$v_f = v_i + at$

when it reached to top point its velocity will become zero

$0 = \sqrt{gS} - g t$

$t = \frac{\sqrt{gS}}{g}$

$t = \sqrt{\frac{S}{g}}$

now the total time of flight will be double of this time as it will take same time to come back on ground

$T = 2\sqrt{\frac{S}{g}}$

Answer 2

Answer:

When body is projected against gravity the initial speed of projection is let say "u" and deceleration due to gravity is "g"

now by the formula of kinematics

v_f^2 - v_i^2 = 2 a svf2−vi2=2as

now we will find the distance traveled by it till it stops

so we will have

0^2 - u^2 = 2(-g)s02−u2=2(−g)s

s = \frac{u^2}{2g}s=2gu2

now we can say that total distance of motion will be double of this because here the distance it will move up will be same as the distance it will come back to ground again.

S = \frac{u^2}{g}S=gu2

from above equation we can find the speed "u"

u = \sqrt{gS}u=gS

now the total time of flight can be found similar way

v_f = v_i + atvf=vi+at

when it reached to top point its velocity will become zero

0 = \sqrt{gS} - g t0=gS−gt

t = \frac{\sqrt{gS}}{g}t=ggS

t = \sqrt{\frac{S}{g}}t=gS

now the total time of flight will be double of this time as it will take same time to come back on ground

T = 2\sqrt{\frac{S}{g}}T=2gS