Question

A stone is thrown vertically upwards with initial velocity u. If g is acceleration due to gravity, then the time after which it returns back to the initial point is
(1) u/g
(2)2u/g
(3)g/u
(4)2g/u​

Answer 1

Answer:

1) u/g

Explanation:

Initial velocity = u = 0

Acceleration = g

Let us consider an object that is projected vertically upwards with initial velocity  u reaching a maximum height h.

Acceleration due to gravity = -g

Following the first equation of motion for a body projected thrown upwards V=u-gt

h=ut-1/2gt²

v²-u²=-2gh

Equations of motion for freely falling body, for the free fall : 

V=gt

h=1/2gt²

v²=2gh

Time of Ascent is the time taken by body thrown up to reach maximum height h 

At maximum height , V=0, thus u=gt1

t1=u/g and maximm height h = u²/2g

Hence, the time of descent,  after reaching maximum height , the body begins to travel downward like a free fall

Thus h = 1/2gt₂²

t₂²=2h/g

t₂=√2h/g

t₂=√2xu²/2g²

t₂=u/g

Therefore, t₁=t₂

The time ascent is equal to time of descent in case of bodies moving under gravity.