menu search
brightness_auto
more_vert
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.
thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike

1 Answer

more_vert
 
verified
Best answer
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=√2u²/2g²
t₂=u/g -----------------equation (10)
∴t₁=t₂
The time ascent is equal to time of descent in case of bodies moving under gravity.
thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike

Related questions

thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 1 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike
1 answer
thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike
1 answer

4.8k questions

4.3k answers

67 comments

388 users

...