Question

michelle and maggie are at baseball practice. michelle throws a ball into the air and when it drops to a height of 5 ft 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)=-16t^2+48t+15.

Need to figure which ball goes higher. Detirmine which girl is standing on a platform & the height of it. And which ball is travelling faster & avg rate of change.

How would I set this up?

Answer 1

cant see the graph for michelles ball.

also for maggie, from the height equation you know that the initial height is 15 ft
unless maggie is a giant, maggie is the one on the platform :)

Answer 2

Thx. Graph attached

Answer 3

the time when ball is at max height can be found using formula
t = -b/2a

for maggie
t = -48/2(-16) = 48/32 = 1.5

Answer 4

Ok

Answer 5

The graph tells you that initial height is 5ft, max height is about 21 ft

Answer 6

OK and how would I calculate which ball is traveling at a faster rate of change on the way up?

Answer 7

that is based on initial velocity
the "b" coefficient in the height function

maggie, its 48 ft/s

michelle, work backwards
from graph max occurs when t=1

t = -b/2a

1 = -b/-32 = b/32 --> b =32 ft/s

maggie's is going faster

Answer 8

Ok. Kinda makes sense. Thanks.

Answer 9

for avg rate of change use how fast it took to reach max height (t)

rate = (max - initial)/t

Answer 10

Ok. Got it on the avg rate!