Question

best answer receives medal and fan :):):):):O


Michelle and Maggie are at baseball practice. Michelle throws a ball into the air and when it drops to a height of 5ft., she hits the ball. The height of the ball is modeled by the graph below where t = time in seconds and h = height of the ball from the ground.
Maggie is throwing a ball into the air and catching it. The height of Maggie’s ball is modeled by the function h(t) = –16t2 + 48t + 15.

Answer 1

Maggie is throwing a ball into the air and catching it. The height of Maggie’s ball is modeled by the function h(t) = –16t2 + 48t + 15.
Part 1. Which ball goes higher in the air, the ball that is hit or the ball that is thrown? Use complete sentences and show all work to explain how you determined the height that each ball reaches.
Part 2. Determine which girl is likely to be standing on a raised platform. Use complete sentences to explain how you determine which girl is on the platform and then determine the height of the platform.
Part 3. Which ball is traveling at a faster average rate of change on the way up? Use complete sentences to explain how you determined the interval at which the height of the ball is increasing and the average rate of change.

Answer 2

where is the graph?

Answer 3

@surjithayer

Answer 4

@dumbsearch2

Answer 5

@poopsiedoodle

Answer 6

for Maggie
h(t)=-16t^2+48t+15\[=-16(t^2-3t+\frac{ 9 }{4 }-\frac{ 9 }{4 })+15=-16\left( t-\frac{ 3 }{ 2 } \right)^2+16*\frac{ 9 }{4 }+15\]
\[=-16\left( t-\frac{ 3 }{2 } \right)^2+51\]
maximum height is 51 and is attained at t=3/2

Answer 7

Mitchel's ball reaches to a height of 22.5 only

Answer 8

okay now what about part b?

Answer 9

b. for Maggie put t=0 and find h(0)
for Mitchel is standing on the ground and hits the ball at height of 5

Answer 10

have you calculated h(0),just plug t=0

Answer 11

\[h(0)=-16(0)^2+48*0+15=?\]

Answer 12

where are you?

Answer 13

im trying to calculate h(0)=

Answer 14

im getting a lil confussed though

Answer 15

it is 15
so Maggie is on a raised platform and the height of platform is 15

Answer 16

ooo okay I see

Answer 17

part 3?

Answer 18

c.for Mitchel when t=o=0, h=5
when t=1,h=22.5
\[Rate~of~change=\frac{ 22.5-5 }{1-0 }=17.5ft/\sec.\]
for Maggie,\[when~t=0,h=15~ft,when~t=\frac{ 3 }{ 2 }\sec.,h=51 ft\]
\[Average~Rate~of~change=\frac{ 51-15 }{\frac{ 3 }{ 2 }-0 }=36*\frac{ 2 }{3 }=24ft/\sec\]
so Maggie's ball is rising at a faster average rate.

Answer 19

can you explain the steps?

Answer 20

just note the height of ball from the graph when t=0 and t=1
difference is change in height.
speed=(change in height)/time

Answer 21

thanks :)

Answer 22

yw