1. Sydney is tossing a softball into the air with an underhand motion. The distance of the ball above her hand at any times given by the formula: h(t)=64t-16t^2.where h(t) is the height of the ball (in feet) and t is time (in seconds).
a. At what times,t, is the ball in her hands?
b. At what time, t is the ball at its maximum height?
c. How high h, is the ball at its maximum height?
To answer these questions, we need to analyze the given formula, which represents the height of the ball above Sydney's hand as a function of time. Let's go through each question step-by-step:
a. At what times, t, is the ball in her hands?
To determine when the ball is in Sydney's hands, we need to find the time(s) when its height, h(t), is equal to zero. In the given formula, h(t) = 64t - 16t^2. To find the times when h(t) = 0, we set the equation equal to zero: 64t - 16t^2 = 0.
Next, we can factor out common terms to simplify the equation: t(64 - 16t) = 0.
Now, we can solve the equation by setting each factor equal to zero:
t = 0 (since multiplying by 0 would result in zero height) or
64 - 16t = 0 (by setting the second factor equal to zero).
Solving the second equation, we have:
64 - 16t = 0
16t = 64
t = 4.
Therefore, the ball is in her hands at two different times: t = 0 and t = 4 seconds.
b. At what time, t, is the ball at its maximum height?
To find the time when the ball reaches its maximum height, we need to determine the vertex of the parabolic function. The formula h(t) = 64t - 16t^2 represents a downward-opening parabola (because of the negative coefficient -16t^2). The vertex of this parabola gives us the maximum height.
The x-coordinate of the vertex can be found using the formula: t = -b/(2a), where a and b are the coefficients of the quadratic equation. In our case, a = -16 and b = 64.
Using the formula, we have: t = -64/(2*(-16)) = -64/(-32) = 2.
Therefore, the ball reaches its maximum height at 2 seconds.
c. How high, h, is the ball at its maximum height?
To determine the height of the ball at its maximum point, we substitute the value of t = 2 seconds into the height function: h(2) = 64(2) - 16(2)^2.
Simplifying the equation, we have: h(2) = 128 - 16 * 4 = 128 - 64 = 64 feet.
Therefore, the ball reaches a maximum height of 64 feet.
(a) find t when h=0 (just solve for t, as usual)
as with all quadratics, the vertex is midway between the roots, at t = -b/2a = -64/-32 = 2