Respuesta :
Using the vertex of a quadratic equation, it is found that:
- It takes 3.27 seconds for the ball to reach it's maximum height.
- It reaches a height of 72.24 feet.
- It takes 7.11 seconds for the ball to hit the ground.
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- A quadratic equation is defined by:
[tex]y = ax^2 + bx + c[/tex]
- It's vertex is:
[tex](x_v, y_v) = (-\frac{b}{2a}, -\frac{b^2 - 4ac}{4a})[/tex]
- If a < 0, the vertex is a maximum point.
The height of a ball after t seconds is given by, considering the gravity:
[tex]h(t) = -4.9t^2 + v(0)t + h(0)[/tex]
- v(0) is the initial velocity.
- h(0) is the initial height.
In this problem:
- Initial height of 20 ft, thus [tex]h(0) = 20[/tex]
- Thrown with a velocity of 32 ft/sec, thus [tex]v(0) = 32[/tex].
The equation is:
[tex]h(t) = -4.9t^2 + 32t + 20[/tex]
Which is a quadratic equation with coefficients [tex]a = -4.9, b = 32, c = 20[/tex].
The vertex is:
[tex]t_v = -\frac{32}{2(-4.9)} = 3.27[/tex]
[tex]h_v = -\frac{32^2 - 4(-4.9)(20)}{4(-4.9)} = -\frac{1416}{4(-4.9)} = 72.24[/tex]
It takes 3.27 seconds for the ball to reach it's maximum height.
It reaches a height of 72.24 feet.
It hits the ground when [tex]h(t) = 0[/tex].
Solving the quadratic equation, the positive root is:
[tex]t = \frac{-32 - \sqrt{1416}}{2*(-4.9)} = 7.11[/tex]
It takes 7.11 seconds for the ball to hit the ground.
A similar problem is given at https://brainly.com/question/24713268
