Mathematics College

Patricia serves the volleyabll to terri with an upward velocity of 19.5 ft/s . The ball is 4.5 feet above the ground when she strikes it. How long does terri have to react before the volleyball hits the ground ? Round your answer to gwo decimal places

Answers

Answer 1

Terri have to react 1.42 seconds before the volleyball hits the ground.

What are quadratic equations?

Quadratics can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. It is also called quadratic equations. The general form of the quadratic equation is:

[tex]\text{ax}^2 + \text{bx} + \text{c} = 0[/tex]

Given data:

Velocity [tex](v_0)[/tex] = 19.5 ft/sHeight [tex](h_0)[/tex] = 4.5 ft

The height can be modeled by a quadratic equation.

[tex]h(t)=-16t^2+v_0t+h_0[/tex]

Where h is the height and t is the time.

[tex]h(t)=-16t^2+19.5t+4.5[/tex]

[tex]-16t^2+19.5t+4.5=0[/tex]

[tex]a = -16, b = 19.5, c = 4.5[/tex]

It looks like a quadratic equation. we can solve it by quadratic formula.

[tex]\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]

[tex]\rightarrow t=\dfrac{-19.5\pm\sqrt{(-19.5)^2-4\times(-16)(4.5)} }{2(-16)}[/tex]

[tex]\rightarrow t=\dfrac{-19.5\pm\sqrt{380.25+288} }{-32}[/tex]

[tex]\rightarrow t=\dfrac{-19.5\pm25.851 }{-32}[/tex]

[tex]\rightarrow t=\dfrac{-19.5-25.851 }{-32}, \ t=\dfrac{-19.5+25.851 }{-32}[/tex]

[tex]\rightarrow t=1.42, \ t=-0.20[/tex]

Time cannot be in negative. So neglect t = –0.235.

Hence, Terri have to react 1.42 seconds before the volleyball hits the ground.

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Answer 2
Final answer:

Using the physics concept of projectile motion and inputting the given values into the appropriate equation, we can determine the time it takes for the volleyball to hit the ground after being served

Explanation:

This question is a classic use of physics, more specifically, the concept of projectile motion. Here, the volleyball can be conceived as a projectile. When Patricia serves the ball upward, the ball will first ascend and then descend due to gravity.

Let's use the following equation which is a version of kinematic equations to solve this problem, adjusting for the fact that we're dealing with an initial height of 4.5 ft and an ending height of 0 ft (when the ball hits the ground). The equation y = yo + vot - 0.5gt² , where:

y is the final vertical position (which we'll take to be 0),yo is the initial vertical position (in this case, the 4.5 feet above the ground),vo is the initial vertical velocity, t is the time (which we're trying to find), andg is the acceleration due to gravity, with the value approximately 32.2 feet per second squared.

Setting y=0, yo=4.5 feet, vo=19.5 feet/second, and g=32.2 feet/second², and plug these values into the equation, we'll get a quadratic equation in the form of 0 = 4.5 + 19.5t - 16.1t². Solve that equation for t to find the time it takes for the ball to hit the ground.

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