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# maximum height

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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

### 2+0 Answers

#1
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(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
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-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