How can we prove that when an object is thrown up and it comes down, the time of ascent is equal to the time of descent?

Let the initial velocity of throwing be u

thus time taken to reach the toppest point be T

thus using the first equation of motion:

T = u/g

distance travelled:

H = uT - 1/2gT²

H = 1/2u²/g

 

now when the ball coms down, the time taken to travel H distance:

u²/(2g) = 1/2gt²

thus t = u/g

hence proved