Prove that for a vertically thrown time of ascent is equal to time of descent

Let the velocity at the time of throwing be u

The time for the velocity to become zero at the highest point of flight is given by first equation of motion

Thus, t₁ = u/g

Height attained, h = u²/2g ----- (uisng the second equation of motion)

Now at the time the time of descent, the time taken for the ball to come down will be:

t₂² = 2h/g ----- (using the third equation of motion)

→ t₂ = (u/g) ---(Since, h = u²/2g)

 

Thus, t₁ = t₂