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

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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)

Initial velocity=u=0
g=g
V=gt----------(4)
h=$$1\over2$$gt²-------(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
$$t_1=\tfrac{u}{g}$$ ----------(7)
Maximm height h=$$u²\over2g$$ ----------(8)

Time of descent : After reaching maximum height , the body begins to travel downward like free fall
so equation (5) h=$$1\over2$$gt₂²
t₂²=$$2h\over g$$
$$t₂=\sqrt{\tfrac{2h}{g}}$$
but from equation (9)
t₂=$$\sqrt{\tfrac{2xu²}{2g²}}$$
t₂=u/g -----------------equation (10)
∴t₁=t₂
The time ascent is equal to time of descent in case of bodies moving under gravity

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