Question

write a quadratic function to model the vertical motion for each situation, given ( h(t) = -16t^2 + v_0t + h_0 ). find the maximum height. initial vertical velocity: 56 ft/s, initial height: 0 ft. the function is ( h(t) = -16t^2 + 56t + 0 ). the maximum height is (\boxed{56}) ft.

Explanation:

Step1: Identify the quadratic function

The general formula for vertical motion is \( h(t) = -16t^2 + v_0t + h_0 \), where \( v_0 \) is the initial vertical velocity and \( h_0 \) is the initial height. Given \( v_0 = 56 \) ft/s and \( h_0 = 0 \) ft, we substitute these values into the formula. So the function is \( h(t) = -16t^2 + 56t + 0 \), which simplifies to \( h(t) = -16t^2 + 56t \).

Step2: Find the time at maximum height

For a quadratic function in the form \( y = ax^2 + bx + c \), the time (or x - value) at which the maximum (or minimum) occurs is given by \( t = -\frac{b}{2a} \). In our function \( h(t) = -16t^2 + 56t \), \( a=- 16 \) and \( b = 56 \).
So, \( t=-\frac{56}{2\times(-16)}=-\frac{56}{-32}=\frac{56}{32}=\frac{7}{4}=1.75 \) seconds.

Step3: Find the maximum height

Now we substitute \( t = 1.75 \) back into the height function \( h(t) \).
\( h(1.75)=-16\times(1.75)^2+56\times(1.75) \)
First, calculate \( (1.75)^2 = 3.0625 \)
Then, \( -16\times3.0625=-49 \)
And \( 56\times1.75 = 98 \)
Then \( h(1.75)=-49 + 98=49 \) feet.

Answer:

The function is \( h(t)=-16t^{2}+56t \) and the maximum height is \( 49 \) ft.