High school maths

Question

Suppose Doug kicks a soccer ball straight up into the air with an initial velocity of 96 feet per second. The function h(t)=-16t^2 +96t gives the height, in feet, of the soccer ball after t seconds? What is the maximum height of the soccer ball? 

Answer 1

Since you put Algebra 1 as the topic, I'm not sure you are in calculus. So in Algebra we learn that this is a parabola that opens downward. So we can use the fact that the X coordinate of the vertex is -b/2a. So in this case t=-96/((2)(-16)) or t=3. At 3 second, the soccer ball reaches its maximum height. To find the height, plug it into the equation and you have -16(9)+96(3)=144ft

Answer 2

Are you currently in calculus? You can take the derivative of the function -> 0=-32*t+96, and then solve for t. This gives you the time value when the ball is at the maximum height. Plug that number back into the original equation to find the answer.

Answer 3

If you had some kinematics in your courses, you might recognize this to be a constant acceleration problem with an equation of motion of the type

 

h(t) = (1/2)at² + v₀t + h₀

 

a = -32 ft/s²

 

v₀ = 96 ft/s

 

h₀ = 0 ft

 

Under constant acceleration, 

 

v(t) = v₀ + at = 96 - 32t (ft/s)

 

At maximum height, v has dropped to 0.

 

96 - 32t = 0

 

t = 96/32 s = 3 s

 

Then h_max = h(3) = [-16(3)² + 96(3)] ft = 144 ft