Question

Question below:

Answer 1

I need help with 58,59,60,61...

It seems like I messed up my calculations.

Answer 2

For 57,I got the first part but couldn't solve for the second part.

Answer 3

57 - just use vo = 0 and plug in the time to find v \[\large v = v_0 + at\]\[\large v = 0 + (-10)t\]

58 - write equations for the position of each stone, using \[\large \Delta y= v_o t +\frac{ 1 }{ 2 }a t ^2\]First stone:\[\large \Delta y= 0* t +\frac{ 1 }{ 2 }(-10) t ^2\]Second stone\[\large \Delta y= 100 t +\frac{ 1 }{ 2 }(-10) t ^2\](not sure if the velocity is really 100; post a bigger picture)
Set equations equal, solve for t.
Plug t back in to find height.

Answer 4

yes the velocity is 100.

Answer 5

-5t^2 = 100t - 5t^2?

Answer 6

100t = 300
t = 3

Answer 7

No 59;

using the equation for constant acceleration v^2 = u^2 + 2as

u - = 10ms-1, a = -10 mx-2 and displacement s = -65

v^2 = 100 + 2*-10*-65

Answer 8

v is the velocity as it hits the bottom

Answer 9

37.41?

Answer 10

What about 60 and 61?

Answer 11

yea i get 37.41 which isnt there - close to 37.14 thogh...

Answer 12

Still trying to figure out what I've done wrong in 58 since it's not making sense.

60 - I'm pretty sure the height on the Moon is just six times the height on the Earth. \[\large v^2 = v_o^2 + 2 a \Delta y\]since initial velocity is the same, the only change is to a, so yes, the height will be six times the Earth height.

61 - Use one of the earlier equations and solve for t \[\large -12 = 8t +\frac{ 1 }{ 2 }(-10)t^2\]

Answer 13

I didnt get #60...

Answer 14

There isn't much to get. Acceleration and height are inversely related; that's all. 1/6th the acceleration means 6 times the height.

Answer 15

1/6th the acceleration means 6 times the height.

is this some theorem or..?

Answer 16

No. It comes from solving for y\[\large v^2 = v_o^2 + 2 a \Delta y\]\[\large 0^2 = v_o^2 + 2 a \Delta y\]\[\large -v_o^2 = 2 a \Delta y\]\[\large -\frac{ v_o^2 }{ 2 a }= \Delta y\]vo is a constant, so... what happens to y if acceleration is multiplied by 1/6

Answer 17

Oh,yes,

In #60
I got :
5t^2 - 8t - 12 = 0

Answer 18

Are there any questions left you need help with? If not, then please close the question

Answer 19

I need some clarification on #55

Answer 20

55 \[\large -500 = 0t + \frac{ 1 }{ 2 }(-10)t^2\]

Answer 21

yep

Answer 22

10?

Answer 23

Another thing to learn/memorize in 55 - time to reach max height = time to fall back to starting height. Sixty percent of the time, it works every time.

58 - still can't figure this out for sure.

Answer 24

in #61
5t^2 - 8t - 12 = 0

i have to use the qudratic formula righT?

Answer 25

Probably yes.

Answer 26

a = 5
b = -8
c = -12
right?

Answer 27

Yes.

Answer 28

i got 2.54 s...but thats not an option..

Answer 29

58 is unsolvable.

Think about it - when the upward projected stone reaches its max height - its speed is zero. At that time, the downward stone has now gained downward velocity... so how is the stone at the top now supposed to catch up? They both accelerate at the same rate.

Answer 30

Well,
Thats a point.

Answer 31

It'd be like dropping a stone from the top of a building, while another one is thrown downward, from halfway up the building. How the hell is the top once supposed to catch up?

Answer 32

@welshfella no wonder 58 wasn't making sense. It's nonsense.

Unless the projected stone is projected from the ground??!

Answer 33

Okay,lets ASSUME its projected from the ground..

Answer 34

yes Agent you are right.

Answer 35

If it is, which it should bloody well say so, let's use\[\large y_f = y_i + v_i t + \frac{ 1 }{ 2 }at^2\]Use y=0 as the ground level, positive is upwards.
First stone:
\[\large y_f = 300 + 0t +\frac{ 1 }{ 2 }(-10)t^2\]Second stone:\[\large y_f = 0 + 100t +\frac{ 1 }{ 2 }(-10)t^2\]Now you can solve for t, then solve for y.

Answer 36

So t=3, using elimination, then plug in for y_f, the final height.

Answer 37

I have to subtract this value from 300 ,right?

Answer 38

so the equation for the stone projected from the ground is

s = 100t - 5t^2

s is distance above the ground at time t

the distance the second stone fell during this time is 300 - s

300 - s = 5t^2

solve these system of equations

Answer 39

No. y=0 is ground height. I assume they want height above ground.

Answer 40

yea s is the height above the ground

Answer 41

@welshfella i was answering "I have to subtract this value from 300 ,right?" btw

Answer 42

OK sorry

Answer 43

anyways
300 - s = 5t^2
s = 300 - 5t^2
also
s = 100t - 5t^2

equate the last 2 and solve for t

Answer 44

t = 3s

Answer 45

so the height must be 225 m..?

Answer 46

height = 300 - 5(3)^2

Answer 47

Isn't it 255m? But i'm not surprised if it's wrong on your question \[\large y_f = 300 +\frac{ 1 }{ 2 }(-10)(3)^2 = 255\]

Answer 48

Oh no the question has it right

Answer 49

lol

Answer 50

we are all burning out xD

Answer 51

well its given 255.9 m which is the value you would get if you used 9.8 ms-2 for gravity

Answer 52

yea I'm exhausted

Answer 53

We're stressed out.

Answer 54

Thanks for the help @welshfella and @agent0smith :)

Answer 55

yw

I've got a cold as well so I'm finished for today!!!

Answer 56

2.52 looks right to me (for no. 61) .