The height from which the ball was released, given the final velocity is closest to 412.46m.
Given the data in the question;
Since, the ball was released from rest.
First we find the time taken for the ball to reach its final velocity
From the First Equation of Motion:
[tex]v = u + at[/tex]
Where v is the final velocity, u is the initial velocity, t is the time and a is acceleration( Since the ball is under gravity, a = g = 9.81m/s² )
[tex]v = u + gt[/tex]
So, we make t the subject of the formula and substitute in our values
[tex]t = \frac{v-u}{g}[/tex]
[tex]t = \frac{90m/s - 0m/s}{9.81m/s^2}[/tex]
[tex]t = \frac{90m/s}{9.81m/s^2} \\\\t = 9.17s[/tex]
To determine the height from which the ball was released.
We use the Second Equation of Motion:
[tex]s = ut + \frac{1}{2}at^2[/tex]
Where u is the initial velocity, t is the time, a is acceleration( Since the ball is under gravity, a = g = 9.81m/s² ) and s is the distance or height.
So, substitute in our values
[tex]h = [ 0m/s\ *\ 9.17s ] + [ \frac{1}{2}\ * 9.81m/s^2\ *\ (9.17s)^2 ]\\\\h = \frac{1}{2}\ * 9.81m/s^2\ *\ 84.0889s^2\\\\h = 412.46 m[/tex]
Therefore, the height from which the ball was released, given the final velocity is closest to 412.46m.
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