Question

If a ball is thrown directly upward with a velocity of 40 ft/s, its heigh in feet after t seconds is given by y=40t-16t^2. What is the maximum height attained by the ball?

I am completely lost on this problem.

Answer 1

\[f(x)=y =ax^2+bx+c\]is the general form for a quadratic.
it has a maximum if a<0 and a minimum if a>0.

the max/min occurs when \[x=\frac{ -b }{ 2a }\]
and the max/min value is \[f \left( \frac{ -b }{ 2a } \right)\]

for your problem, \[f(x) = y = -16t^2 +40t\]
a = -16 and b = 40
since a < 0 you will have a maximum. it occurs when \[x = \frac{ -b }{ 2a }=\frac{ -(40) }{ 2\cdot (-16) }=\frac{ -40 }{ -32 }=\frac{ 5 }{ 4 }\]

so plug in...

\[f \left( \frac{ 5 }{ 4 } \right)=-16\left( \frac{ 5 }{ 4 } \right)^2 + 40\left( \frac{ 5 }{ 4 } \right)=-25+50=25\]

Answer 2

sorry, used x when it should have been t.

Answer 3

Thank you so much!!!

Answer 4

does that make sense?

Answer 5

yes it does! Thanks!! :)