+0  
 
0
59
2
avatar

a ball is thrown vertically upward. after T seconds, its height H (in feet) is given by the function h(t)=44t-16t^2. What is the maximum height the ball will reach?

 Mar 3, 2021
 #1
avatar+352 
0

(I'm not very good at functions... but I think this is correct)

h(t)=44t-16t^2

t=44/32=11/8

h(t)=44(11/8)-16(11/8)^2=121/2-121/4=$\boxed{\frac{121}4}$

 Mar 3, 2021
 #2
avatar
0

-16t^2 + 44t      = h       max h will occur at t = - b/2a =  -  44/(2*-16) =  1.375 sec  

     Use this value of t to calc height

 

-16 ( 1.375)^2 + 44 ( 1.375) = 30.25 ft max

 Mar 3, 2021

33 Online Users

avatar