Jason kicked a soccer ball that was lying on the ground. It was in the air for 3 seconds before it hit the ground again. While the soccer ball was in the air it reached a maximum height of 30 ft. The height of the ball while in the air is the function of time.

Answer:

a) The domain is 0 < t < 3

The range is 0 < s < 30

b) The x-intercept are the times at the start and return of the ball to the ground

The y-intercept, is the starting height of the ball = 0

c) The axis of symmetry is the line t = 1.5, and the maximum function is the value of the height at the vertex of the parabola, they represent the turning point of the function

d) The rate of change of the function is therefore given by the derivative of function which is;

ds/dt  = d(14.715×t - 1/2×9.81×t²)dt = 14.714 - 9.81×t which is the velocity of the ball at time t

∴ ds/dt = 14.714 - 9.81×t

Which is the velocity of the ball at time t

Step-by-step explanation:

The duration the soccer ball was in the air = 3 s

The maximum height reached by the soccer ball = 30 ft.

Considering the vertical motion of the ball, we have;

v = u - gt

s = u·t + 1/2·g·t²

v² = u² - 2×g×s

Where;

v = The final velocity of the ball = 0 m/s

u = The initial velocity of the ball

g = The acceleration due to gravity = 9.81 m/s²

s = The height the ball reaches = 30 ft

Given that the time to maximum height = Half the total time it was in the air, we have;

v = u - gt

v = 0 at maximum height

u = gt = 9.81 × 3/2 = 14.715 m/s

The vertical velocity = 14.715 m/s

Therefore, we have;

s ≈ 14.715×t - 1/2×9.81×t²

a) The domain is 0 < t < 3

The range is 0 < s < 30

b) The x and y-intercept are;

x-intercept, s = 0, gives;

0 ≈ 14.715×t - 1/2×9.81×t²

From which we have, t = 0 or t = 3, which are the times at the start and return of the ball to the ground

The x-intercept are the times at the start and return of the ball to the ground

The y-intercept, t = 0, gives;

s = 14.715×t - 1/2×9.81×t² = 14.715×0 - 1/2×9.81×0 = 0

Therefore, at the time t = 0 the ball is lying on the ground, height = 0

∴ The y-intercept, is the starting height of the ball = 0

c) The axis of symmetry is the line t = 1.5, and the maximum function is the value of the height at the vertex of the parabola

They represent the turning point of the function

d) The rate of change of the function is therefore given by the derivative of function which is;

ds/dt  = d(14.715×t - 1/2×9.81×t²)dt = 14.714 - 9.81×t

∴ ds/dt = 14.714 - 9.81×t

Which is the velocity of the ball at time t.

HELP PLS I HABE BEEN STARING AT THUS QUESTION FOR SO LONG​

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