Question 42672
I am supposing that you are at ground level.
s = -16t^2 + v0t + s0
a) What is the function that describes this problem?
Parabolic due to the square of a variable.
b) The ball will be how high above the ground after 1 second?
s = -16t^2 + v0t + s0
s = -16t^2 + 64t
s = -16(1)^2 + 64(1) = -16 + 64 = 48 feet
c) How long will it take to hit the ground?
s = -16t^2 + 64t
{{{a=-16}}} and {{{b=64}}} and {{{c=0}}}
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (-64 +- sqrt( 64^2-4*(-16)*(0) ))/(2*(-16)) }}} 
{{{x = (-64 +- sqrt( 64^2 ))/(-32) }}}
{{{x = (-64 +- 64)/(-32) }}}
{{{x = (-64 + 64)/(-32) }}} or {{{x = (-64 - 64)/(-32) }}}
{{{x = 0 }}} or {{{x = 4 }}}
After four seconds, the ball will hit the ground.
{{{ graph( 600, 600, -10, 10, -10, 65, -16x^2 + 64x) }}}
d) What is the maximum height of the ball? What time will the maximum height be attained?
You would need to know the vertex.
s = -16t^2 + 64t
Vertex: ((-b/2a),f(x)) = ((-64/-32),f(x)) = (2,64)
The maximum height is 64 feet after 2 seconds.