Question

A ball is released from rest at a certain height give its final velocity is 90 meters per second the higher the Bal was released from is closest to

Answer 1

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.

• Initial velocity; $u = 0m/s$

• Final velocity; $v = 90 m/s$

• The acceleration due to gravity; $g = 9.81m/s^2$

First we find the time taken for the ball to reach its final velocity

From the First Equation of Motion:

$v = u + at$

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 )

$v = u + gt$

So, we make t the subject of the formula and substitute in our values

$t = \frac{v-u}{g}$

$t = \frac{90m/s - 0m/s}{9.81m/s^2}$

$t = \frac{90m/s}{9.81m/s^2} \\\\t = 9.17s$

To determine the height from which the ball was released.

We use the Second Equation of Motion:

$s = ut + \frac{1}{2}at^2$

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

$h = [ 0m/s\ *\ 9.17s ] + [ \frac{1}{2}\ * 9.81m/s^2\ *\ (9.17s)^2 ]\\\\h = \frac{1}{2}\ * 9.81m/s^2\ *\ 84.0889s^2\\\\h = 412.46 m$

Therefore, the height from which the ball was released, given the final velocity is closest to 412.46m.

Learn more; brainly.com/question/13275688