Question

The height of a rock at time t which is thrown vertically on the height 46 feet is given
by the formula s(t) = −t
2 + 20t+ 44. What is the maximum height of the rock? When
does it hit the ground? What is the impact speed?

Answer 1

yes it is

Answer 2

im just stuck on when the rock his the ground and the impact i figured out the height

Answer 3

@jfernandes

Testing...

Answer 4

Not sure I'm wondering if we can just take the derivative of S(t) and set it equal to zero.



\[\frac{ dS }{ dt } = -2t+20 \]

\[\frac{ dS }{ dt }_{0} \]

\[-2t+20 = 0 \]

\[\frac{ -20 }{ -2 } = 10 = t \]

I feel that when the velocity is zero the will reach it's maximum height.

so we plug this back into the original equation.

\[s(2) = -(2)^{2}+20(2)+44 = 80ft\]

When the rock hit's the ground the position will be zero so

\[s(t) = -t ^{2}+20t+44 = 0 \]

i'm thinking we can apply this formula

\[-b \pm (b ^{2}-4ac)^{.5}/2a\]

Answer 5

so far the maximum height is 144 and i checked to see if it was right and it was

Answer 6

oh wait.. i plugged in the wrong number

Answer 7

so i the derivative at 0 t = 10
then i plugged it back in and got -(10)^2 + 20(10) + 44
= 144 which is the max height

Answer 8

so now im wondering how to i find te value of t when the rock hits the ground

Answer 9

\[-(10)^{2} + 20(10)+44\]
-100+200 = 100+44 = 144

@reeses Yeah I plugged in 2 instead of 10

Answer 10

when the rock hits the ground @reeses the height will be zero, vertical height

Answer 11

the acceleration is constant, the flight is symmetrical, so time to max height, is same as time down

Answer 12

-b +/- (b^2-4ac)^0.5/2a

((20)^2-4(-1)(44))^0.5

-20-19.6 = -40/2(-1) = 20 seconds

Answer 13

is that 46 the initial height, i didnt see that?

Answer 14

i'm thinking that to find the velocity we would plug t back into the equation for velocity.

dS/dt = -2t+20

t = 20

-2(20)+20 = -20 m/s i'm guessing the negative sign means it's downward

@DanJS I did not see that, oh... yeah .. the initial height was 46ft but @reeses said the answer for height was correct I'm guessing?

Answer 15

sweet! i got it now so it was -(t-22)(t+2)=0
t = 22
t = -2 which -2 isnt right
so t = 22 is right and its the right answer

Answer 16

now how about the impact

Answer 17

Yeah, in the prob, the position function at time 0 is s(t) = 44, but it says starts at 46,

Answer 18

i know its throwing me off

Answer 19

@reeses if you found the time at impact , I think you can plug it back into the equation, the derivative dS/dt which gives velocity

Answer 20

yes that is for when it hits the ground, it has to fall that extra 44 ft on the way down, with a initial velocity the same as the initial velocity it was thrown upwards

Answer 21

well the time of impact to the ground is 22

Answer 22

yeah so 10 to top , 10 back to the h=44, and must be 2 more sec down

Answer 23

if you use the initial velocity and the accel and height, should get that time

Answer 24

how would you set it up?

Answer 25

so i looked up the answer and its 4 feet/sec i have no idea how thats the answer

Answer 26

s(t) = -t^2 + 20t + 44 [ft]

v(t) = -2t + 20 [ft/s]

a(t) = -2 [ft/s^2]

--------------------------------

Answer 27

The initial height s(0) = 44 ft, thrown up with an initial velocity of v(0) = 20 ft/s

undergoes a constant acceleration of -2 ft/s^2

Answer 28

kimematic equation to get time from 20 ft/s to 0 ft/s with constant accel of -2 is


10 seconds

Answer 29

but how did the text get 4 feet/sec

Answer 30

s(10) = max height

time to fall from that max height to h=0 must be 22 s

Answer 31

sorry 12s

Answer 32

could it be a typo?

Answer 33

the ground inpact speed will be greater (neg direction) than the initial velocity because it is accelerating downwards longer than deceleration to zero speed,