You throw an orange out a window at a height of 12.00 meters upwards at an angle of 32° to the horizontal at a velocity of 3.5 m/s. What is the magnitude and direction of the velocity vector after .50 seconds?

First, determine the vertical velocity after .50 seconds. In order to do this, you must use the equation v_{y} = v_{0y} + gt. We know that v_{0y }= v_{0}sinθ = 1.85 m/s from previous problems. Therefore, v_{y} = 1.85 + (-9.81) (.50) = -3.06 m/s. Next, you must find the horizontal velocity vector. We know that v_{x} = v_{0}cosθ = 2.97 m/s.

Now that you have both the horizontal and vertical components, you must use the Pythagorean theorem to solve for the overall velocity vector (vector addition) .

Therefore, at t = .50 s, v_{0} = √ (v_{x}^{2} + v_{y}^{2}) = √ (2.97^{2} + 3.06^{2}) = 4.3 m/s

Lastly, we must determine the direction of the velocity vector. In order to do this, you must use the formula: θ = tan^{-1} (v_{y}/v_{x}) . Plugging in our values, we get θ = 46°.