To watch the video A ball is projected directly upward from an initial height of 200 feet with an initial velocity of 120 feet per second. After how many seconds does the ball reach its
maximum height? What is this height?

The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.

Step-by-step explanation:

Let us take the point of projection of the ball as origin of the coordinate system, the upward direction as positive and down direction as negative.

Initial velocity u with which the ball is projected upwards = + 120 ft/s

Uniform acceleration a acting on the ball is to acceleration due to gravity = - 32 ft/s²

The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.

Using the formula:

v² - u² = 2 a h,

where

u = initial velocity of the ball = +120 ft/s

v = final velocity of the ball at the highest point = 0 ft/s

a = uniform acceleration acting on the ball = -32 ft/s²

h = height attained

Substituting the values we get;

0² - 120² = 2 × (- 32) h

=> h = 120²/2 × 32 = 225 feet

The height of the ball from the ground at its highest point = 225 feet + 12 feet = 237 feet.

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743

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step-by-step explanation: