From a height of 3 meters above ground level, a rock with mass m = 4 kg is hurled into the air with an initial velocity of 10 m/sec. Let y denote the rock's height, in meters above the ground, at time t seconds after it is hurled. In addition to a gravitational force with Fg = mg with g = 9.8 m/sec2, the moving rock experiences air resistance directly proportional to the velocity, y', that is, a resistive force Fr = cy', with c = 0.02. The model my'' = −(Fg + Fr) describes this situation.
To the nearest 0.01 seconds, when does the rock achieve its maximum height?
To the nearest 0.1 meters, what is the maximum height that it achieves?

No bc a lot of things r still theories n unproven

Wooden , walls ,doors ,support

Ibelieve the answer is b

    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