Answer:
The acceleration is [tex]a= 2.4 \ m/s^2[/tex]
Explanation:
From the question we are told that
The distance covered is [tex]d = 87 \ m[/tex]
The time taken is [tex]t = 8.6 \ s[/tex]
Time taken reach the bottom is [tex]t_b = 1 \ s[/tex]
According to the equation of motion
[tex]S = ut + \frac{1}{2} at^2[/tex]
since the ball started at rest u = 0 m/s
substituting values
[tex]87 = 0 + \frac{1}{2} * a * (8.6)^2[/tex]
=> [tex]a = \frac{2 * 87}{8.6^2}[/tex]
=> [tex]a= 2.4 \ m/s^2[/tex]
