A ball is thrown up into air from the ground and the ball's velocity is modeled by the equation v(t)=-(1)/(4)x^(2)-4cos(pi x)+8. At what t value does the ball reach its maximum?

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A ball is thrown up into air from the ground and the ball's velocity is modeled by the equation v(t)=-(1)/(4)x^(2)-4cos(pi x)+8.  At what t value does the ball reach its maximum?

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Identify Velocity Equation: Identify the given velocity equation for the ball. The given velocity equation is v(t) = -(1/4)x^2 - 4cos(pi x) + 8. We need to find the value of t at which the velocity is zero, as this will be the point where the ball changes direction and starts to fall back down, indicating the maximum height.Set Equation Equal to Zero: Set the velocity equation equal to zero to find the critical points. 0 = -(1/4)x^2 - 4cos(pi x) + 8 This is a transcendental equation because it involves both polynomial and trigonometric terms. We will need to use numerical methods or graphing techniques to solve for x.Use Graphing Calculator: Use a graphing calculator or software to plot the velocity equation and find the t value where the velocity is zero. Since the equation is transcendental, we cannot solve it algebraically for an exact solution. Instead, we will approximate the solution using a graphing tool.Observe Graph for Maximum Height: Observe the graph to determine the t value where the velocity crosses the t-axis, which indicates the maximum height. By graphing the function, we can see that the velocity crosses the t-axis at approximately t = 2 seconds. This is the point where the ball reaches its maximum height.

A ball is thrown up into air from the ground and the ball's velocity is modeled by the equation $v(t)=-\frac{1}{4}x^{2}-4\cos(\pi x)+8$ . $\newline$ At what $t$ value does the ball reach its maximum?

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A ball is thrown up into air from the ground and the ball's velocity is modeled by the equation $v(t)=-\frac{1}{4}x^{2}-4\cos(\pi x)+8$ . $\newline$ At what $t$ value does the ball reach its maximum?