Question

At a particular baseball stadium the outfield wall is 310 feet away from home plate and the wall is 6 feet tall. A player hit the ball when it was 3 feet off of the ground with an initial upward velocity of 64 feet per second and an initial outward velocity of 77 feet per second.
Write the function h(x) that gives the height of the ball after time (t).

Answer 1

i am really rusty at this - i know i could do these at one time. I'll Be back in half hour with the answer if u still waiting.

Answer 2

O.k cool thanks! Anyone else know how to do this?

Answer 3

ok you use one of the constant acceleration equation take the acceleration due to gravity to be 32 ft/s/s

h(x) = ut + 0.5 g t^2
where u = initial upward velocity, t=time and g = acceleration due gravity

so h(x) = 64t - 0.5*32t^2
h(x) = 64t - 16t^2

the negative is due to gravity acting in opposite direction to motion of the ball

oh sorry - i missed out the initial 3 ft so its:

h(x) = 64t - 16t^2 + 3

Answer 4

Thank you so much! So based off of this, how long does the ball stay in the air?

Answer 5

ok u can use the same equation to work this
now h = 0 so

0 = 64t - 16t^2 + 3
solving this t use the formula:

t = -64 + sqrt(64^2 - 4 * -16 * 3 / -32

this gives t = 4.05

Answer 6

seconds that is

Answer 7

Wow. You and amistre64 were really close! He got 4.04 seconds. I understand your explanation though. Thanks again for the help jimmyrep!

Answer 8

ok

Answer 9

plz click good answer to close question when u are done

Answer 10

Oh wait! Another thought:
How far will the ball travel? This is based off of the time (4.04 seconds).

Answer 11

How long did the ball take to return to a height of 6 feet?

Answer 12

AND FINALLY:
How far did the ball travel before it returned to a height of 6 feet?

Answer 13

ok - i'// take 4.04 seconds - igot 4.046 and rounded it up - but whatever

the horiznontal distance the ball will travel is given by

v = distance / time where v = initial outward velocity
distance = v * t = 77 * 4.04 = 311 feet

Answer 14

Hmm. O.k got it. Now I just need help on the last two questions.

Answer 15

the distance the ball travels is the outward velocity times 4.04 right?

Answer 16

it started at 3 ft so at 6ft it will be 3 ft higher and so we can use

3 = 64t - 16t^2

solving this the time is 3.95 seconds

Answer 17

77 ft per sec times 4.04 secs = distance traversed

Answer 18

yep

Answer 19

Wait so how far will the ball travel?

Answer 20

it lands abt 311 feet away; how long is the path that it takes thru the air tho?

Answer 21

distance travelled when at height 6ft =

3.95 * 77 = 304.2 ft

Answer 22

How long did the ball take to return to a height of 6 feet?

Answer 23

distance through the air? that will be a parabola right?

Answer 24

"How far did the ball travel before it returned to a height of 6 feet?" I interpret this to mean that the ball is on the downward arc ....

Answer 25

yes, the distance along the parabola; but I dont think that was the right interp for the question :)

Answer 26

right - thats what I thought

Answer 27

(-16t^2 +64t +3) - 6 = 0 tells us the times at h=6ft right?

Answer 28

yes - i think thats right

Answer 29

i get the upward part is abt; .05 secs

the downward is abt; 3.95 secs

Answer 30

yes

Answer 31

How long did the ball take to return to a height of 6 feet? The answer to this would be which one though?

Answer 32

to return means to come back to; it can only come back to 6ft if its coming back down again; which is the farthest one away and the longest time to get there

Answer 33

Oh so downward

Answer 34

So now back to distance. How far did the ball travel before it returned to a height of 6 feet?

Answer 35

jimmyrep said:

distance travelled when at height 6ft = 3.95 * 77 = 304.2 ft

Answer 36

i'm struggling a bit with my memory of doing these
the horizontal component of the flight is constant - no acceleration so if the time to get to 6ft above the ground is 3.95 sec then
distance = time x speed
= 3.95 x 77 = 304.2 ft

i'm about 95 % sure thats correct

Answer 37

Are either of you good at tessellations?

Answer 38

it correct; we assume that there is no loss of acceleration; or that the loss is neglible given the short amount of time

Answer 39

So this answer is the same answer for the question of How far will the ball travel?

Answer 40

4.04 * 77 is not the same answer as:

3.95 * 77 .....

Answer 41

one measure total distance traveled while airborne; the other measures total distance traveled as it arcs thru 6 ft

Answer 42

So 3.95*77 is how we would get the answer to how far the ball traveled?

Answer 43

Which would be 304.15 feet?

Answer 44

yes - thats the horizontal distance travelled

Answer 45

O.k and the ball traveled 304.2 feet before it returned to a height of 6 feet? Just making sure.

Answer 46

well i think that this distance is the horizontal distance travelled from the player to a point directly below the ball when it is descending at a height of 6 feet. Thats how I have interpreted the question.

Answer 47

Hmm. O.k. Well THANK YOU SO MUCH GUYS FOR YOUR HELP! I couldn't have done it without YOU.

Answer 48

Jimmy, are you familiar with tessellations?

Answer 49

sorry mathkid i'm not

Answer 50

its all good

Answer 51

What was the maximum height the ball reached?

Answer 52

are u still there ? i had to go away a while

Answer 53

Yes, I just need to find out what was the maximum height the ball reached.

Answer 54

Would it be 512 feet?

Answer 55

v^2 = u^2 + 2 xg x h
at maxm height the speed v = 0

0 = 64^2 - 2 *32 * h
h = 4096 / 64 = 64

add initial 3 ft

height = 67 ft

Answer 56

Thanks! :D

Answer 57

JIMMYREP, i posted another problem... if you wouldnt mind :/

Answer 58

ok