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
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
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