By applying the quadratic formula and discriminant of the quadratic formula, we find that the maximum height of the ball is equal to 75.926 meters.

<h3>How to determine the maximum height of the ball</h3>

Herein we have a quadratic equation that models the height of a ball in time and the maximum height represents the vertex of the parabola, hence we must use the quadratic formula for the following expression:

- 4.8 · t² + 19.9 · t + (55.3 - h) = 0

The height of the ball is a maximum when the discriminant is equal to zero:

19.9² - 4 · (- 4.8) · (55.3 - h) = 0

396.01 + 19.2 · (55.3 - h) = 0

19.2 · (55.3 - h) = -396.01

55.3 - h = -20.626

h = 55.3 + 20.626

h = 75.926 m

By applying the quadratic formula and discriminant of the quadratic formula, we find that the maximum height of the ball is equal to 75.926 meters.

To learn more on quadratic equations: brainly.com/question/17177510

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6 0

2 years ago

Only find the vertex and axis of symmetry

lara [203]

Answer:

vertex (5,2)

axis of symmetry: x=5

Step-by-step explanation:

vertex (h,k)

y = a(x - h)² + k

f(x)=(x-5)²+2 a = 1 h = 5 k = 2

vertex (5 , 2)

The axis of symmetry always passes through the vertex of the parabola. The x -coordinate of the vertex is the equation of the axis of symmetry of the parabola.

x = 5

5 0

3 years ago

Forces of 9 lbs and 13 lbs act at a 38º angle to each other. Find the magnitude of the resultant force and the angle that the re

fiasKO [112]

Answer: $R=20.84\ lb\quad 22.57^{\circ},15.43^{\circ}$

Step-by-step explanation:

Given

Two forces of 9 and 13 lbs acts $38^{\circ}$ angle to each other

The resultant of the two forces is given by

$\Rightarrow R=\sqrt{a^2+b^2+2ab\cos \theta}$

Insert the values

$\Rightarrow R=\sqrt{9^2+13^2+2(9)(13)\cos 38^{\circ}}\\\Rightarrow R=\sqrt{81+169+184.394}\\\Rightarrow R=\sqrt{434.394}\\\Rightarrow R=20.84\ lb$

Resultant makes an angle of

$\Rightarrow \alpha=\tan^{-1}\left( \dfrac{b\sin \theta}{a+b\cos \theta}\right)\\\\\text{Considering 9 lb force along the x-axis}\\\\\Rightarrow \alpha =\tan^{-1}\left( \dfrac{13\sin 38^{\circ}}{9+13\cos 38^{\circ}}\right)\\\\\Rightarrow \alpha =\tan^{-1}(\dfrac{8}{19.244})\\\\\Rightarrow \alpha=22.57^{\circ}$

So, the resultant makes an angle of $22.57^{\circ}$ with 9 lb force

Angle made with 13 lb force is $38^{\circ}-22.57^{\circ}=15.43^{\circ}$

7 0

3 years ago

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