Question

A particle is moving along a projectile path where the initial height is 96 feet with an initial speed of 16 feet per second. What is the maximum height of the particle?

Answer 1

Assuming the initial speed is upwards, the position function is \[s(t)=-16t^2+16t+96\]

Answer 2

To find the time of the maximum height, let t=-b/2a=-16/-32=1/2 sec

Answer 3

Put that into the function to get the maximum height.

Answer 4

wait so whats the answer

Answer 5

Would you work out under his intruction?

Answer 6

um ok...

Answer 7

i just don't really get it

Answer 8

We're right here for you :)

Answer 9

Plug 1/2 into the function in the top post of the thread, and see what you get.

Answer 10

t=1/2
s(t)=−16t2+16t+96
=> s ( 1/2) = ...

Answer 11

oh okay so s(1/2)= -16(1/2)^2 + 16(1/2) + 96
s(1/2) = 8^2 + 8 + 96
s(1/2)= 168
?????

Answer 12

i have no idea

Answer 13

s (1/2) = 6 ( -t² + t + 6)

Answer 14

where did 6 come from

Answer 15

I just want to make it easier for you!

Answer 16

Your order of operations is wrong.
\[s(1/2) = -16(1/2)^2 + 16(1/2) + 96=-16(1/4)+8+96=-4+8+96=100\]

Answer 17

I mean 16 :")

Answer 18

I don't think six divides 16 evenly...

Answer 19

OK.

Answer 20

o ok so its 100?

Answer 21

@nicole1217 all you do is plug the time in, the value S obtain is the height!

Answer 22

Yep! high five :)

Answer 23

yay! Thanks :)

Answer 24

@nicole1217
Just want to emphasize the key idea that Ani. mentioned at first.
"To find the time of the maximum height, let t=-b/2a=-16/-32=1/2 sec

Answer 25

yeah i got it now