Question

attachment below

Answer 1

https://answers.yahoo.com/question/

Answer 2

@phi

Answer 3

I assume the z-axis is the height, and z=0 is ground level.

the kid is at the top of the tower at t=0.
plug in z=0 to find her height (and therefore the height of the tower)

Answer 4

how far did you get with this one?

Answer 5

i didn't know where to start. the teacher have not even went over these type of problems

Answer 6

you should be able to a and b. give it a shot.

Answer 7

is a = 120 and b = 0 ?

Answer 8

a is ok.

b should be the time to get to the bottom
i.e. solve for t when z=0

Answer 9

120/5 = t ?

Answer 10

120 - 5t= 0
120 = 5t
120/5 = t
yes

Answer 11

for c, do i need to derive ?

Answer 12

yes, we find the velocity in each dimension
\[\text{ Let } \dot{x}=\frac{dx}{dt} \]
to find the total velocity we use
\[ v = \sqrt{ \dot{x}^2 +\dot{y}^2+\dot{z}^2}\]

Answer 13

-25sint + 25cost - 5
\[\sqrt{(-5)^{2}+25^{2}} \]

Answer 14

you can find the velocity vector
\[ \vec{v}= < \dot{x}, \dot{y},\dot{z}>\]
and then find the magnitude of that vector.

notice it has *three* components.

Answer 15

so \[\frac{ 25-5 }{ \sqrt {(-5)^2 +25^2} }\]

Answer 16

what is that ?

Answer 17

velocity vector < -25sint , 25cost ,- 5>

speed when t = 0 \[|v(t)| = \sqrt{25^{2} + (-5)^2}\]

Answer 18

ok,

but they are asking for the speed at time t (not just at t=0)
however, notice
(-25 sin t)^2 + (25 cos t)^2 + 25
is
25^2 sin^2 t + 25^2 cos^2 t + 25
factor out 25^2 from the first two terms
25^2 ( sin^2 t+ cos^2 t) + 25

sin^2 t + cos^2 t = 1 , and you get
25^2 + 25

in other words, the velocity is *always*
\[ \sqrt{650} \]
for any time t

Answer 19

so c is
\[ 5 \sqrt{26} \approx 25.5 \]

Answer 20

okay, Thanks
so for d would i derive the velocity vector, and find the magnitude of the acceleration right?

Answer 21

yes

Answer 22

the z term will disappear
(it's constant for the velocity)
so it's simpler than the v vector.

Answer 23

would it be \[\sqrt{25^2} = 25\]

Answer 24

yes.

Answer 25

Thank you