Example Question - maximum height ball
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Calculating Time to Reach Maximum Height of a Ball
To solve for the time it takes for the ball to reach its maximum height, we can look at the equation of motion that is given: \[ h = -16t^2 + 32t + 5 \] In this equation, \(h\) represents the height of the ball in feet after \( t \) seconds. The maximum height is reached at the vertex of the parabola represented by this quadratic equation. The time at which the maximum height is achieved can be found by using the formula \( t = -\frac{b}{2a} \), where \( a \) is the coefficient of \( t^2 \) and \( b \) is the coefficient of \( t \). In this equation, \( a = -16 \) and \( b = 32 \). Plugging these values into the formula gives us: \[ t = -\frac{32}{2(-16)} \] \[ t = -\frac{32}{-32} \] \[ t = 1 \] So, it takes 1 second for the ball to reach its maximum height.
Calculating Time to Reach Maximum Height of Ball
The question is asking for the time it takes for the ball to reach its maximum height when thrown upward. The given equation for the height h in feet at any time t in seconds is: h(t) = -16t^2 + 32t + 5 To find the time when the ball reaches its maximum height, we need to determine the vertex of the parabolic function since the coefficient of the t^2 term is negative, indicating that the parabola is concave down and the vertex will give the maximum height. For a quadratic equation in the form of ax^2 + bx + c, the x-coordinate of the vertex (which gives us the time in our case) can be found using the formula: t = -b/(2a) In our equation, a = -16 and b = 32. So plugging these values into our formula: t = -32 / (2 * -16) t = -32 / -32 t = 1 Therefore, it takes 1 second for the ball to reach its maximum height.
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