its fine lucky this online hw site lets you submit multiple times, so i still got it right

Well, you should be doing the homework yourself first or second anyway, so it won't matter if my assistance has an error in it.

okay ive been working on this one for quite a while

While riding on an elevator descending with a constant speed of 3.6 m/s, you accidentally drop a book from under your arm.
How long does it take for the book to reach the elevator floor, 1.0 m below your arm?

i tryed using the equation y=y0+v0t+.5at^2
y = 0
y0 = 1m
t = ?
a = g = 9.8m/s
t = ?

now after solving this, i keep getting nonreal or negative answers... what am i doing wrong?

This one is a trick. Because the elevator is moving at a constant velocity, you ignore the fact that it's moving down, and look at it as a fixed reference frame.

Your variables are almost correct, but a = NEGATIVE 9.8m/s^2

0 = 1 + 0t -4.9t^2
-1=-4.9t^2
sqrt(1/4.9) = t
t = sqrt(0.204) = .45 seconds

alright, one more thing. ive done the rest of my homework but this problem has me stumped. we are dealing with vectors now. so here the problem

As two boats approach the marina, the velocity of boat 1 relative to boat 2 is 2.10 m/s in a direction 40.0° east of north. If boat 1 has a velocity that is 0.725 m/s due north, what is the velocity (magnitude and direction) of boat 2?


a)________________ m/s
b__________________Degrees south of west

OK, I'll write vectors as V and scalars as v. Boat one has velocity V1, boat two is V2. We'll first break Boat 1's velocity down into components:

v1x = 0.725cos(90) = 0 m/s
v1y = 0.725sin(90) = 0.725 m/s

We probably could have figured that you without the math, but the formalism is nice. Now, we know the following:

v1x = v2x + 2.10cos(50) = v2x + 1.35 m/s //note that "40° E of N" is 50° N of E
v1y = v2y + 2.10sin(50) = v2y + 1.61 m/s

Therefore, we can easily do the subtraction:

v2x = v1x - 1.35 = 0 - 1.35 = -1.35 m/s
v2y = v1y - 1.61 = 0.725 - 1.61 = -0.885 m/s

We can then turn that back into magnitude-angle form:

V2 = sqrt(-1.35^2 + -0.885^2) @ tan^-1(-0.885/-1.35)°
V2 = 1.614 @ 33+180° = 1.614 m/s @ 213°

Also known as 1.614 m/s going 33° south of east.

amazing Kerm! i dont know how you do it. thanks so much

Well, if you look at my solution, it's just simple addition plus steps of conversion between magnitude-angle and vector-component forms. Was that answer correct?

Kerm told me to post this here.

Projectile range help? http://math.pastebin.com/qxsj7s7n `-`

So, we'll have to deal with some vector stuff. Things we know right off the bat:

- angle of firing = O [for theta] **be with consistent degrees/radians**
- initial speed V0 = 30.0 m/s
- initial x speed v0x = 30cos(O)
- initial y speed v0y = 30sin(O)
- desired x final xf=R=20.0m
- desired y final yf=0.0m

We'll set up our usual equations:

vx = v0x + ax*t
vy = v0y + ay*t

We know that there is no horizontal acceleration, so vx = v0x at all times. We know that the vertical acceleration is due to gravity, so we can state:

vy = v0y + at

where a = -9.8m/s^2. Next, we'll visit our position kinematics equations:

x = x0 + v0x*t + 0.5*ax*t^2
y = y0 + v0y*t + 0.5*ay*t^2

As we already said, these can get simplified to the following:

x = x0 + v0x*t
y = y0 + v0y*t + 0.5*a*t^2

Now we can start plugging in some values, like our xf and yf!

xf = 20 = x0 + v0x*t
20 = 0 + v0x*t = v0x*t

yf = 0 = y0 + v0y*t + 0.5*a*t^2
0 = 0 + v0y*t -4.9*t^2
0 = v0y*t - 4.9*t^2

Well, it looks like we'll need a value for t, doesn't it? Obviously there will be two different possible values for t, which sound like a classic quadratic equation to me. We have two unknowns, essentially: t and O, the flight time and the firing angle. We also don't know v0x and v0y, but those have fixed values based on O. We have four equations now that we can use:

20 = v0x*t
0 = v0y*t - 4.9*t^2
v0x = 30cos(O)
v0y = 30sin(O)

Might as well do some substitution:

20 = 30cos(O)*t
0 = 30sin(O)*t - 4.9*t^2

Look at that, two equations and two unknowns! Go us. We'll apply the quadratic equation to the second equation:

a = -4.9
b = 30sin(O)
c = 0
t = [-30sin(O) +/- sqrt(900*sin^2(O) - 0)]/-9.8

Yeesh, that looks pretty unwieldy. Let's do some simplification.

t = [30sin(O) +/- abs(30sin(O))]/9.8
t = 3.061sin(O) +/- 3.061sin(O)
t= 6.122sin(O) or t=0

Obviously only the first one will be valid. We've painted ourselves into this corner:

20 = 30cos(O)*t
t= 6.122sin(O)

Wait, but is this bad? No! 6.122sin(O) seconds after liftoff, we'll hit the ground, and exactly at t=0, we're also at the ground. Finally, all we have to do is solve:

20/(30cos(O)) = 6.122sin(O)
(2/3)/6.122 = sin(O)cos(O)

Because I worked on and off on this for well over an hour, need to go to sleep, and have my own problems to worry about, I simply plug this into my friendly neighborhood TI-83+ and hit solve:

0.1089 = sin(O)cos(O)
O = [0.1098 rad, 1.461 rad] = 6.29deg, 83.7deg

You can easily figure out the travel time yourself. I'm tired, so you should really check my work.

A projectile is shot from the edge of a cliff 205 m above ground level with an initial speed of vo=125 m/s at an angle of 35.0° with the horizontal, as shown in the figure.



Determine the distance X of of point P from the base of the vertical cliff.

At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Assume the positive directions are upward and to the right.)

Rhombus: Although i am not in physics i think you can model it in your calc using parametric equations

X1T=(55/90)*125*T
Y1T=-9.8T^2+(35/90)*125*T+205

T is number of seconds

Window settings:
Xmin:0
Xmax:1000
Ymin:0
Ymax:300
Tmin:0
Tmax:26

Correct me if I am wrong

I'm really hesitant to solve this. I screwed up my sleep schedule for the week solving Keith's problem, and I've heard nary a Thank You for my efforts. I suggest you just use trig to figure out the first part, knowing that vx0 = vx = vxf = v0cos(theta), and vy0 = v0sin(theta), and vy varies according to vy = vy0 + a*t where a=-9.8m/s.

Don't worry Kerm, i understand completely.

v0x = v0cos(theta) = 102.4m/s
v0y = v0sin(theta) = 71.7m/s

x = x0 + v0x*t = v0x*t = 102.4*t
y = y0 + v0y*t + 0.5*a*t^2 = h + 71.7*t - 4.9*t^2 = 205 + 71*t - 4.9*t^2

Of course, we know we want to find when y = 0:

0 = 205 + 71*t - 4.9*t^2

Now we apply the quadratic formula:

a = -4.9
b = 71.7
c = 205

t = {-2.45, 17.08}

So it takes 17.08 seconds for the projectile to hit the ground. That makes things easy:

x = 102.4*17.08 = 1749.18m

Now for part 2. We know that v0x = 102.4m/s still, because we have no acceleration or wind resistance in the x direction. Now we solve for vy:

vy = v0y + a*t = 71.7 + (-9.8)*(17.08) = -95.7 m/s

Excuuuuuuuuse me, Martian. I thanked you on IRC and promised to some day send you thank-you calcs when I have money. What more do you want?! My soul!?

.... *ahem* Thank you for the help Kerm, however your solution did confuse me towards the end and I actually found a simpler solution that works, albeit probably through bad assumptions `-`;

It's not your fault, I'm just crabby in general these days. You didn't force me to answer you. I'm curious what the better solution you found was?

I guess you'll be needing these then
*Keith gives Kerm a cane and a lawn to stand on whilst waving it.


Anyway, my solution:

R = (v0^2 / g)sin(2θ)
sin^-1(Rg / v0^2) / 2 = θ
θ = 6.28927232 degrees ~= 6.29 degrees (3 sf)
90 – θ = 83.71072768 ~= 83.7 degrees (3 sf)

I don't really know if 90 - θ is considered valid in calculating the greatest possible angle, but the math worked out if your answers were correct

Yup, 90-theta is indeed the correct way to do it, and will generally hold true for most of these problems.

Woo! Is that because the least and greatest possible firing angles are naturally complementary?