Acar in a movie moving at 20 m/s flies off a cliff that is 74.3m tall. how long is the car in the air for and how far does the car travel before it hits the ground below?

We use the equation S=½gt² where S=vertical distance=74.3m, g, acceleration of gravity=9.8 m/s/s. So t=√(2S/g)=3.884 seconds. So the car is in the air for about 3.894 secs.
It’s horizontal distance is vt=20×3.894=77.88m.

This is what i found on google “what happens to the bond angle when you add or remove an electron domain? the bond angle decreases as you add domains 5. can you force the atoms into new configurations by pushing atoms around? note: we can think of a bond or a lone pair of electrons as a “domain” of electrons.” i hope this you

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s