Verified answer

A ball is thrown directly upward with an initial velocity of 10 m/s. If the ball starts at an initial height of 2.6 m, how long is the ball in the air? Ignore air drag,

Final velocity

Y = Yo + Vo * t – ½ * g * t^2

Y – Yo = Vo * t – ½ * g * t^2

Y = final position = ground = 0 m

Yo = initial position = +2.6m

Y – Yo is the displacement of the ball = 0 – 2.6 = -2.6

-2.6 = 10 * t – ½ * 9.8 * t^2

-2.6 = 10 * t – 4.9 * t^2

Add 4.9 * t^2 – 10 * t to both sides

4.9 * t^2 – 10 * t – 2.6 = 0

This is a quadratic equation

Time = [10 ± (10^2 – 4 * 4.9 * -2.6)^0.5] ÷ (2 * 4.9)

Time = [10 ± (150.96)^0.5] ÷ 9.8

Time = [10 + (150.96)^0.5] ÷ 9.8 = 2.27 seconds

OR

Time = [10 – (150.96)^0.5] ÷ 9.8 = -0.233 seconds

Time = 2.27 seconds