A water-balloon launcher with mass 2 kg fires a 0.75 kg balloon with a velocity of 14 m/s to the west. What is the recoil velocity of the launcher?
What is the answer

Correct V₂ = 5.25 m/s

Explanation:

Given:

m₁ = 0.75 kg mass of balloon, V₁ = 14 m/s the velocity of the balloon

m₂ = 2 kg mass of water balloon launcher

The law of impulse sustainability applies for system water balloon launcher - balloon which means:

J₁ = m₁ · V₁   Impulse of the balloon after launching

J₂ = m₂ · V₂  Impulse of the water balloon launcher

these impulses are equal

J₁ = J₂ ⇒  m₁ · V₁ = m₂ · V₂ ⇒ V₂ = (m₁ · V₁) / m₂

V₂ = (0.75 · 14) / 2 = 5.25 m/s

V₂ = 5.25 m/s

God is with you!!!

vapour or gas assuming atmospheriv pressure

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min