Question

a stone thrown vertically upwards with initial velocity u reaches a height h before coming down show athat the time taken to go up the same as the time taken to come down.

Answer 1

Hello Dear.


To Prove ⇒ Time of Ascent = Time of Descent.

Proof ⇒ 

In First Case,(When the Ball is thrown upwards),

Initial Velocity = u

Final Velocity(v) = 0 
[Since, the ball must stops after reaching the certain heights]

Height to which the ball reaches(S) = h

Acceleration(a) = -g  [Since, the ball is thrown against the gravity)
= -10 m/s²

Now, Using the Equations of the Motion,

v² - u² = 2aS
(0)² - u² = 2(-10)h
u² = 20h
⇒ u = (√20h)

Now,
∵ v - u = at
∴ 0 - u = (-10)t
u = 10t
⇒ t = u/10
⇒ t = (√20h)/10    -----------eq(i)

Thus, the time of ascent is (√20h)/10 seconds.


In Second Case,(When the balls falls from the certain Height),
Acceleration(a) = g
= 10 m/s²
Initial Velocity(u) = 0
Final Velocity(v) = v

∵ v² - u² = 2aS
∴ v² - (0)² = 2(10)h
⇒ v² = 20h
⇒ v = (√20h)

Now,
Using the Equation,

  v - u = at.
∴ v - 0 = 10t
⇒ v = 10t
⇒ t = v/10
⇒ t = (√20h)/10 seconds.  ---------eq(ii)

Thus, time of Descent is (√20h)/10 seconds.


We can see that the eq(i) and the eq(ii) are equals, thus,Time of Ascent is equals to the time of Descent.

Hence Proved.


Hope it helps.