Abored kid, 30 meters from a building throws a ball at an angle of 50 degrees with a velocity of 20m/s at the building wall. at what height above the throwing line will the ball hit the wall?

9.1 meters First, determine the vertical and horizontal components of the ball's velocity. So Vhorz = cos(50)*20 m/s = 0.64278761 * 20 m/s = 12.85575219 m/s Vvert = sin(50)*20 m/s = 0.766044443 * 20 m/s = 15.32088886 m/s Now determine how long until the ball hits the building. Do this by dividing the horizontal distance by the horizontal velocity. T = 30 m / 12.85575219 m/s = 2.33358574 s Now the expression for the height of the ball is given by h = VT - 0.5 AT^2 where h = height V = initial vertical velocity T = Time A = acceleration due to gravity. Substituting known values. h = (15.32088886 m/s)(2.33358574 s) - 0.5 * 9.8 m/s^2(2.33358574 s)^2 h = 35.75260778 m - 4.9 m/s^2 * 5.445622407 s^2 h = 35.75260778 m - 26.6835498 m h = 9.069057982 m Rounding to 2 significant figures gives 9.1 meters. So the ball will impact the building 9.1 meters above the height at which the ball was thrown.

    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