quadratic equations

Question

a ball is shot from a cannon into the air with an upward velocity of 40 ft/sec.

the eqaution that gives the hieght (h) of the ball at any time (t) is: = -16t^2 + 40ft + 1.5

 

a) find the maximum height attained by the ball

 

b) how long did it take the ball to it the ground ? 

Answer 1

A)

 

To find the maximum height,  we can set the derivative of h equal to zero, since the derivative is the slope of the tangent line and the slope of the line tangent to the maximum is zero.

 

h' = 0

 

-32t + 40 = 0

 

-32t = -40

     t = 1.25

 

 

This value of t is when the ball reaches its maximum height.  Substitute this value of t into the original function to obtain the maximum height.

 

h = -16t² + 40t + 1.5

h = -16(1.25)² + 40(1.25) + 1.5

h = 26.5

 

The maximum height is 26.50 ft.

 

 

B)

 

Set h=0.

 

0 = -16t² + 40t + 1.5

 

Use the quadratic formula.

 

t = (-40 ± √(1600 - 4(-24))) / -32

t = (-40 ± √(1696)) / -32

t = (-40 ± 41.18) / -32

 

t = (-40 + 41.18) / -32           and            t = (-40 - 41.18) / -32

 

t = -0.037                              and            t = 2.537