Question

1. A stone thrown vertically upwards with
initial velocity u reaches a height 'h'
before coming down. Show that the
time taken to go up is same as the time
taken to come down.
​

Answer 1

Answer

Time = √(2s/g)

Given

A stone thrown vertically upwards with  initial velocity u reaches a height 'h'  before coming down

To prove

The  time taken to go up is same as the time  taken to come down

Formula's Applied

→ v = u + at

→ v² - u² = 2as

Solution

Case - 1 : Time taken to go up

Final velocity , v = 0 m/s

Since , finally goes to rest at extreme position

Acceleration due to gravity , a = - g m/s²

Since , thrown against gravity

Apply 1st equation of motion ,

⇒ v = u + at

⇒ (0) = u + (-g)t

⇒ 0 = u - gt

⇒ gt = u

⇒ t = u/g ... (1)

Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (0)² - u² = 2(-g)s

⇒ -u² = -2gs

⇒ u² = 2gs

⇒ u = √(2gs)

Sub. u value in (1) ,

⇒ t = u/g

⇒ t = √(2gs)/g

⇒ t = √(2s/g)

Case - 2 : Time taken to come down

Initial velocity , u = 0 m/s

Since , starts from rest

Acceleration due to gravity , a = g m/s²

Apply 1st equation of motion ,

⇒ v = u + at

⇒ v = (0) + (g)t

⇒ v = gt

⇒ t = v/g ... (2)

Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ v² - (0)² = 2(g)s

⇒ v² - 0 = 2gs

⇒ v = √(2gs)

Sub. v value in (2) ,

⇒ t = v/g

⇒ t = √(2gs)/g

⇒ t = √(2s/g)

In both cases , time , t is same .

Hence proved

So , Time taken to go up is same as the time taken to come down