Question

a stone thrown vertically upwards with initial velocity u reaches the height at before coming down. Show that the time taken to go up is same as the time taken to come down.

Answer 1

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HERE IS THE ANSWER TO YOUR QUESTION.....

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

Acceleration due to gravity=-g

Equation of Motion for a body projected thrown upwards :

V=u-gt----------(1)

h=ut-1/2gt² -----------(2)

v²-u²=-2gh --------(3)

Equations of motion for freely falling body :

for free fall :  

Initial velocity=u=0

g=g

V=gt----------(4)

h=1/2gt²-------(5)

v²=2gh------(6)

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

At maximum height , V=0

Equation (1) turns to u=gt1

t1=u/g  ----------(7)

Maximm height h=u²/2g ----------(8)

Time of descent : After reaching maximum height , the body begins to travel downward like free fall

so equation (5) h=1/2gt₂²

t₂²=2h/g

t₂=√2h/g

but from equation (9)

t₂=√2xu²/2g²

t₂=u/g -----------------equation (10)

∴t₁=t₂

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

Answer 2

Acceleration due to gravity=-g

Equation of Motion for a body projected thrown upwards :

V=u-gt----------(1)

h=ut-1/2gt² -----------(2)

v²-u²=-2gh --------(3)

Equations of motion for freely falling body :

for free fall :

Initial velocity=u=0

g=g

V=gt----------(4)

h=1/2gt²-------(5)

v²=2gh------(6)

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

At maximum height , V=0

Equation (1) turns to u=gt1

t1=u/g ----------(7)

Maximm height h=u²/2g ----------(8)

Time of descent : After reaching maximum height , the body begins to travel downward like free fall

so equation (5) h=1/2gt₂²

t₂²=2h/g

t₂=√2h/g

but from equation (9)

t₂=√2xu²/2g²

t₂=u/g -----------------equation (10)

∴t₁=t₂