Derek kicks a soccer ball off the ground and in the air with an initial velocity of 31 feet per second. Using the formula H(t) = -16t2 + vt + s, what is the maximum height the soccer ball reaches?

Question

jdoe0001 · Answer

well keep in mind that $\bf \text{initial velocity}\
h = -16t^2+v_ot+h_o \text{ , in feet}\
v_o = \textit{initial velocity}\
h_o = \textit{initial height}\
h = \textit{height at "t" seconds}$

jdoe0001 · Answer

so plugging in the data available
it says that the ball got kicked "off the ground", so the initial height =0
the initial velocity is given, 31 ft/s
so our equation will look like $\bf H(t) = -16t2 + 31t + 0$

anonymous · Answer

so solve for t

jdoe0001 · Answer

notice that the equation is a parabola.... well $\bf H(t) = -16t^2 + 31t + 0$
it has a negative leading coefficient, meaning that it's opening downwards like
|dw:1377715444670:dw|

anonymous · Answer

ok. whats a coefficient?

jdoe0001 · Answer

well, no, solving for "t" will give us the "x-intercepts" or when the ball is at the ground
we want to know that maximum of the parabola, that is, the "hump" point
or the vertex point, keep in mind that usually the y-axis represents the Height, the x-axis represents the Seconds

so we want the y-axis value at the vertex point
for a parabola like so, you can always find it's vertex at $\bf \left(-\cfrac{b}{2a}, c-\cfrac{b^2}{4a}\right)$

jdoe0001 · Answer

http://www.mathsisfun.com/definitions/coefficient.html

anonymous · Answer

do you have an answer because i am going to do the math let me know if i got it right?

jdoe0001 · Answer

no I don't, but I can get it from $\bf \left(-\cfrac{b}{2a}, c-\cfrac{b^2}{4a}\right)$   quick enough

jdoe0001 · Answer

the y-axis component is the only one you need from the vertex, the y-axis is the Height at the vertex point