A baseball is thrown into the air with an upward velocity of 25 ft./s. It's height in feet after T seconds can be modeled by the function h(t)=-16t^2 +25t+5. how long will it take the ball to reach it's maximum height? What is the balls maximum height?
how about complete the square to find the vertex?
16 t^2 -25 t = -h + 5
t^2 -1.5625 t = -(1/16)h + .3125
(t - .78125)^2 = -(1/16 )h + .3125 +.61035
(t - .78125)^2 = -(1/16 )h + .923
(t - .78125)^2 = -(1/16 )(h - 14.8)
t = .781
h = 14.8
or
0 = 25 - 32 t
t = .781
h = 5 + 25(.781) - 16(.781)^2 = 14.8
To find the time it takes for the ball to reach its maximum height, we need to determine the vertex of the function. The formula for the vertex of a quadratic function in the form h(t) = at^2 + bt + c is given by t = -b / (2a).
In this case, the quadratic function is h(t) = -16t^2 + 25t + 5.
Comparing it to the general form, we have a = -16, b = 25, and c = 5.
Plugging in these values into the formula for the vertex, we get:
t = -25 / (2 * -16)
t = -25 / -32
t = 0.78125 seconds
Therefore, it will take approximately 0.78125 seconds for the ball to reach its maximum height.
To find the maximum height, we can substitute the value of t into the function to get h(0.78125):
h(t) = -16(0.78125)^2 + 25(0.78125) + 5
h(t) = -10.1875 + 19.53125 + 5
h(t) = 14.34375 feet
Therefore, the ball's maximum height is approximately 14.34375 feet.
To find the time it takes for the ball to reach its maximum height, we need to determine the highest point on the graph of the function. The maximum height occurs at the vertex of the quadratic function.
The vertex of a quadratic function in the form \(f(x) = ax^2 + bx + c\) is given by the formula \(x = -\frac{b}{2a}\).
In our case, the function representing the height of the ball is \(h(t) = -16t^2 + 25t + 5\). So, a = -16 and b = 25.
Using the vertex formula, we can find the time it takes for the ball to reach its maximum height:
\(t = -\frac{b}{2a} = -\frac{25}{2(-16)} = \frac{25}{32}\) seconds.
Therefore, it will take the ball \(\frac{25}{32}\) seconds to reach its maximum height.
To find the maximum height, we substitute this time back into the function:
\(h\left(\frac{25}{32}\right) = -16\left(\frac{25}{32}\right)^2 + 25\left(\frac{25}{32}\right) + 5\)
Evaluating this expression gives us the maximum height of the ball.