Answer:
1.24 seconds
Explanation:
To answer the question we will use the following equation:
[tex]y_f=y_i+v_it+\frac{1}{2}at^2[/tex]Where:
yf is the final height of the ball, so it is 0m
yi is the initial height of the ball, yi = 20 m
vi is the initial velocity which is vi = -10 m/s because it is downward
a is the acceleration due to gravity a = 9.8 m/s²
t is the time.
So, replacing the values, we get:
[tex]\begin{gathered} 0=20-10t+\frac{1}{2}(-9.8)t^2 \\ 0=20-10t-4.9t^2 \\ -4.9t^2-10t+20=0 \\ \text{Then} \\ a=-4.9 \\ b=-10 \\ c=20 \end{gathered}[/tex]To find the time, we need to solve this equation. So, using the quadratic formula, we get:
[tex]\begin{gathered} t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t=\frac{-(-10)\pm\sqrt[]{(-10)^2_{}-4(-4.9)(20)}}{2(-4.9)} \\ t=\frac{10\pm\sqrt[]{100+392}}{-9.8} \\ t=\frac{10\pm22.18}{-9.8} \end{gathered}[/tex]Therefore, the values for t are:
[tex]\begin{gathered} t=\frac{10+22.19}{-9.8}=-3.28 \\ t=\frac{10-22.19}{-9.8}=1.24 \end{gathered}[/tex]Since t = -3.28 doesn't have sense here, the answer is 1.24 seconds.
The ball reaches the ground after 1.24 seconds.
