Aflagpole is perpendicular to the horizontal but iis on a slope that rises 10 degrees from the horizontal. the pole casts a 43-foot shadow down the slope and angle of elevation of the sun measured from the slope is 36 degrees. how tall is the pole?

36 ft. The flagpole and the shadow create a triangle with the angle at the base of the flagpole being 100 degrees. The angle at the end of the shadow being 36 degrees and the angle at the top of the flagpole is 180-100-36 = 44 degrees. We know the length of the shadow is 43 feet, and using the law of sines, we get sin(44)/43 = sin(36)/X where X is the height of the flagpole. So let's solve for X sin(44)/43 = sin(36)/X Xsin(44)/43 = sin(36) Xsin(44) = 43sin(36) X = 43sin(36)/sin(44) X = 43 * 0.587785252/0.69465837 X = 36.38445446 Rounding to 2 significant figures, gives us a height of 36 ft.

    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