# Question aaab8

Jul 4, 2015

Here's an alternative method you can use to solve this problem.

#### Explanation:

You can determine the distance covered by your ball in its thirds seconds of flight by subtracting what it travelled in 2 seconds from what it travelled in 3 seconds.

${h}_{\text{third second}} = {h}_{3} - {h}_{2}$

Two seconds after its launch, the object covered

${h}_{2} = {v}_{0} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

${h}_{2} = 4 \text{m"/cancel("s") * 2cancel("s") + 1/2 * 10"m"/cancel("s"^2) * 2^2cancel("s"^2) = "28 m}$

Three seconds after its launch, the ball covered

${h}_{3} = 4 \text{m"/cancel("s") * 3cancel("s") + 1/2 * 10"m"/cancel("s"^2) * 3^2cancel("s"^2) = "57 m}$

Therefore, the distance it covered in its third seconds of flight is

h_"third second" = 57 - 28 = color(green)("29 m")#