Question

A ball is thrown across a playing field from a height of h = 6 ft above the ground at an angle of 45° to the horizontal at the speed of 20 ft/s. It can be deduced from physical principles that the path of the ball is modeled by the functiony = −32/((20)^2)x^2 + x + 6 where x is the distance in feet that the ball has traveled horizontally Find the maximum height attained by the ball. (Round your answer to three decimal places.)

(b) Find the horizontal distance the ball has traveled when it hits the ground. (Round your answer to one decimal place.

Answer 1

for max height differentiate...and put dy/dx = 0
find values of x..put em into the original eqn

Answer 2

its not working for me can u plz try the solution for me

Answer 3

20^2 in the denominator?

Answer 4

yeah

Answer 5

dy/dx = -64x/400 + 1 = 0
x = 400/64
= 25/4

Answer 6

now put in x=25/4 into ur original eqn

Answer 7

yeah i kept the 6.25 in the solution
but i am not getting the right answer

Answer 8

9.125

Answer 9

is it?

Answer 10

IS IT?

Answer 11

it works now

Answer 12

I think I kept the wrong answer I got 15.375

Answer 13

do u know the second part?

Answer 14

well do calculations correctly

Answer 15

what class r u in?

Answer 16

college grade

Answer 17

ohk now the second part

Answer 18

when the ball falls down what is its height?

Answer 19

o

Answer 20

so set y = 0

Answer 21

and find x

Answer 22

wow u made it very easy for me

Answer 23

dont sweat it

Answer 24

wht were u thinking of?

Answer 25

finding x value now

Answer 26

found?

Answer 27

nope

Answer 28

iam unable to find

Answer 29

use quadratic formula

Answer 30

am using the same

Answer 31

not getting

Answer 32

am getting it as complex solution

Answer 33

33.875

Answer 34

how did u get it
can u plz sho the steps when am using quadratic functions its giving me complex value

Answer 35

33.875 is wrong answer@ him

Answer 36

16.9375

Answer 37

now right?

Answer 38

how did u get that value can u plz show me the steps?

Answer 39

\[x = (-b \pm \sqrt{b^2 - 4ac})/2a\]

Answer 40

its exact value u got it perfect? I used the same formula but I got in complex

Answer 41

a = -32/400
b= 1
c =6

Answer 42

do it properly

Answer 43

ull get it

Answer 44

made a small mistake

Answer 45

got it now

Answer 46

made wrong coz of my time is running off thanx a lot for your help

Answer 47

no problem
click good answer

Answer 48

If a ball is thrown directly upward with a velocity of 40 ft/s, its height (in feet) after t seconds is given by y = 40t − 16t2. What is the maximum height attained by the ball? (Round your answer to the nearest whole number.)

Answer 49

16t^2 in the above one

Answer 50

25 ft

Answer 51

click on good answer on the right of my name

Answer 52

how did u get it directly?

Answer 53

i calculated in my head

Answer 54

nothing direct about it

Answer 55

did u click?

Answer 56

yeah

Answer 57

can u tell me the steps?

Answer 58

h(t) = 40t - 16t^2
dh/dt = 40 -32t = 0
t = 1.25
put it in the eqn

Answer 59

gimme a medal

Answer 60

i already gave u

Answer 61

doesnt show any..lol

Answer 62

u became my fan

Answer 63

i clicked on the medal is it not?

Answer 64

click on "good answer" beside the medal

Answer 65

njoi the day dude

Answer 66

thanks man

Answer 67

u 2