Respuesta :

Answer:

About 2.19 seconds

Explanation:

[tex]d=v_ot+\dfrac{1}{2}at^2[/tex]

Since the ball is dropped from rest, there is no initial velocity, and you can write the following equation:

[tex]23.5=\dfrac{1}{2}(9.8)t^2 \\\\t^2\approx 4.79 \\\\t\approx 2.19s[/tex]

Hope this helps!

It will attain a speed of 20m/s before touching the ground, but the time taken to reach the ground is 2 seconds. Because the velocity is not uniform but acceleration/gravity is. Therefore, the answer to your question is: 2 seconds.