SOLUTION: The heoght in feet, s, of a ball thrown straight up is given by the function s(t)=-16t^2 + 56t+40, where t is the time in seconds. Find the maximum height reached by the ball.

Algebra ->  Graphs -> SOLUTION: The heoght in feet, s, of a ball thrown straight up is given by the function s(t)=-16t^2 + 56t+40, where t is the time in seconds. Find the maximum height reached by the ball.      Log On


   

Question 168517: The heoght in feet, s, of a ball thrown straight up is given by the function
s(t)=-16t^2 + 56t+40, where t is the time in seconds. Find the maximum height reached by the ball.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The height in feet, s, of a ball thrown straight up is given by the function
s(t)=-16t^2 + 56t+40, where t is the time in seconds. Find the maximum height reached by the ball.
:
This is a quadratic equation, max height will be at axis of symmetry
The formula for this is x = -b/(2a)
:
In this equation: x = t; a =-16; b = 56
t = %28-56%29%2F%282%2A-16%29 = t = %28-56%29%2F%28-32%29
t = +1.75 seconds at max height
:
Find the max height, substitute 1.75 for t in the original equation,
s(t) = -16(1.75^2) + 56(1.75) + 40
s(t) = -16(3/0625) + 98 + 40
s(t) = -49 + 98 + 40
s(t) = 89 ft is max height