Question

time =2usin thita/g how?

Answer 1

Vy= vsintheta where vcostheta is zero so using equation of motion v= u + at Vsintheta = gt. So t= vsintheta/g which is time of ascent and time of descent so total time= 2vsintheta/g. Hope it helps you.

Answer 2

Time is not always necessary to be 2usin∅/g it is only in case of projectile motion .
Then what is projectile motion ?
Motion of object under the effect of gravity is called projectile motion .

Consider an object projected with an angle of ∅ with respect to horizontal with an initial velocity of u

Then rectangular components of
Y - coordinates
initial velocity = uy = usin∅
acceleration = ay = gsin-90 = -g
when projectile hits ground then
displacement is y - axis = Sy = 0
time = t = T

S = ut + at²/2

Sy = uyT + ayT²/2

0 = usin∅T -gT²/2

0 = (2usin∅T - gT²)/2

0 = 2usin∅T - gT²

0 = (2usin∅ - gT )T

0 = 2usin∅ - gT

-2usin∅ = -gt

T = -2usin∅/-g

T = 2usin∅/g