Let h is the height of the ball after t seconds
The acceleration upward = -32 feet/sec.^2
This situation must be represented by a quadratic function
The form of the function is:
[tex]h=ut+\frac{1}{2}at^2+h_0[/tex]u is the initial velocity
a is the acceleration of gravity
t is the time
h0 is the initial height
From the given, the initial height is 1 foot
The acceleration of gravity is a constant value -32 ft/s^2
The initial velocity is unknown
Let us substitute the values given in the function
[tex]\begin{gathered} h=ut+\frac{1}{2}(-32)t^2+1 \\ h=ut-16t^2+1 \end{gathered}[/tex]Let us arrange the terms from greatest power of t
[tex]h=-16t^2+ut+1[/tex]We have only 1 function in the choices similar to our function
[tex]h=-16t^2+25t+1[/tex]The answer is the second choice
