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

Asked by seeni2005 | 7th Sep, 2015, 09:11: PM

Expert Answer:

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

Answered by Yashvanti Jain | 8th Sep, 2015, 10:20: AM