Question

answer this how can and solve
and dont give irrelevant answers​

Answer 1

Time of flight in absence of air :

t0 (t zero) = 2u/g

t0 = 2×12 / 10

t0 = 2.4 s

Time lf flight in presence of air :

a(net) = a(d) + g

Now, T_{a,u} = u/(g + a)

g + a(d) = 12

a(d) = 2 m/s^2

S= 1/2 a't'²

⟶ 6 = 1/2 × (10 - 2) t'²

⟶ 12/8 = t'²

⟶ t' =√(3/2)

⟶ t' = 1.22 s

Difference Δt = [2.4 - (1 + 1.22)]

Δt=0.18s

Hence, difference of time of flights of the ball in presence and absence of air friction is 0.18s