Question

Prove that the time of descend is equal to its ascend of a given object

Answer 1

Refer to the attachment.

Hope you get your answer.

Answer 2

Hiii.....here is your answer.

i) If you launch a projectile with some force,As soon as the force is no longer acting on the object (time,t=0) it's entire system is known.
ii) The force of gravity is the only thing operating on the object which also decrease a tiny amount as the object moves from the earth.
iii) As the object travels,it will eventually come to a stop and then begins the decent.Each gravity will still apply in relation to its height,there are no new force .
iv) The acceleration on decent is due to only the object having pushed is way through the gravity force to get to that height.
V) So it should exactly match how the object traveled from launch,in reverse.

Let's solve a equation.
: During ascent ,
let the initial velocity be u. and final velocity, v =0
we have,
v = u +at
0 = u+(-g)t
0 = u-gt
gt = u
:. t =u/g
Maximum height reached
We have,
v^2= u^2+2as

So,
0 = u^2-2gh
2gh =u^2
h = u^2/2g.

: During descent,
let initial velocity = 0
acceleration due to gravity = -g
Displacement,s = -h

We have,
s = ut+ 1/2 at^2
-h = 0 -gt^2/2
h =gt^2/2
We have,
h = u^2/2g
So,
u^2/2g = gt^2/2
from which we get
t = u/g
: Therefore time of ascent = The time of descent .

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