The ball is thrown vertically upward at a speed of 49 m / s. After what time it will be at a height of h = 102.9 m.

Data: V0 (ball throwing speed) = 49 m / s; h (the height at which the ball will be) = 102.9 m.

Reference values: g (acceleration due to gravity, acceleration of the ball) ≈ 10 m / s2.

The duration of the ball’s motion is determined from the formula: S = h = V0 * t – g * t ^ 2/2.

g * t ^ 2/2 – V0 * t + h = 0.

10 * t ^ 2/2 – 49 * t + 102.9 = 0.

5t ^ 2 – 49t + 102.9 = 0.

D = b ^ 2 – 4ac = 49 ^ 2 – 4 * 5 * 102.9 = 343.

t1 = (-b + √D) / 2a = (49 + √343) / (2 * 5) = 6.75 s.

t2 = (-b – √D) / 2a = (49 – √343) / (2 * 5) = 3.05 s (correct).