Question

Derive This Equation T1 +T2=2U/G

Answer 1

Explanation:

when particle is moving upward direction,

time = t₁

height = h

velocity when it is reached to the maximum altitude and became stationary is 0 and acceleration in upward direction is -g. Thus,

v = u + at₁

0 = u + (-gt₁)

u = gt₁

t₁ = u/g

Distance traveled upward direction:

v² = u² + 2as

0 = u² 2(-g)s

u² = 2gs

s = h = u²/2g

Now we will calculate the time taking during falling,

h = ut + 1/2at²

u = 0

h = 0t + 1/2gt²

h =  1/2gt²

t² = 2h / g

By taking square root,

t = √2h/g

So the time taken by falling object at a distance h is,

t₂ = √2h/g...... (a)

as we calculated above that distance traveled in upward direction is

h = u²/2g

Thus by putting the values in (a)

t₂ =  √2u²/2g/g

t₂ =  √2u²/2g²

t₂ = u/g

t₁ +  t₂= u/g + u/g

t₁ +  t₂=  2u/g

Answer 2

Answer:

Above answer is correct