Se avienta una piedra desde la azotea de un edificio de 50 metros de altura con una trayectoria parabólica. La velocidad inicial es 25 m/s con un ángulo de 45° respecto a la horizontal. Determina lo siguiente: - Velocidad vertical en el punto de la altura máxima

- Velocidad horizontal en el punto de la altura máxima

- Magnitud de la velocidad en el punto de la altura máxima

- Tiempo en el que se alcanza la altura máxima

- Altura máxima

- Tiempo que tarda la piedra en tocar el suelo.

- Distancia horizontal recorrida

- Velocidad vertical justo al momento de tocar el suelo

- Velocidad horizontal justo al momento de tocar el suelo

- Magnitud de la velocidad justo al momento de tocar el suelo

My answer is opptionc, just check with anyone else

    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

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Ithink it's  c. i'm sure