Question 65416
Imagine you are serving a volleyball underhanded.  suppose you hit the ball when it is 3 feet above the ground, and it leaves your hand with an initial upward velocity of 20 feet per second.
general formula for height:  H=-1/2*gt^+vt+h where g is the acceleration due to gravity, and on the earth, equals 32ft/sec^.  v is the initial upward velocity of the object and h is the initial height above the ground.
the question is:  what is the maximum height that the volleyball will reach and how many seconds will it take to reach this height?
h=3, v=20, and g=32
The formula for height:
{{{H=(-1/2)(32)t^2+(20)t+3}}}
{{{H=-16t^2+20t+3}}}
The maximum height is the H value of the vertex.
The formula for finding the t value of the vertex is: {{{t=-b/2a}}} for a quadratic equation in this form: {{{H=at^2+bt+c}}}
a=-16, b=20, and c=3
{{{t=-20/(2(-16))}}}
{{{t=20/32}}}
{{{t=.625}}}
The maximum height H, will happen when t=.625 s.
{{{H=-16(.625)^2+20(.625)+3}}}
{{{H=-16(.390625)+20(.625)+3}}}
{{{H=-6.25+12.5+3}}}
{{{H=9.25}}}
The Maximum height is: 9.25 ft or 9 ft 3 in
Happy Calculating!!!