Aball is thrown vertically upward, which is the positive direction. a little later it returns to its point of release. the ball is in the air for a total time of 8.00 s. what is its initial velocity? neglect air resistance.

Final Velocity = Vf = 0 m/s > (Vf = 0 because ball's speed at its max height is 0) Initial Velocity = Vi = ? Total time (upward & downward) = 8.0 seconds * Time upward = 4 seconds & ( As time for ball upward & downward is equal ) * Time downward = 4 seconds.. Gravitational Acceleration = g = -9.8 m/s² Use Equation; Vf = Vi - gt 0 = Vi - 9.8 * 4 0 = Vi - 39.2 39.2 = Vi => Vi = Initial Velocity = 39.2 m/s

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way