Physics Help

[JustAnAverageGuy]

Originally posted by: DrPizza
ohhh, so it lands on top of the fence? Then you've gotta add the radius of the baseball to the distance and only consider the lower of the two velocities.

I know you're being sarcastic, but I don't think they'd make a Physics 1 problem that hard

 

[DrPizza]

I'm trying to think of an easier way to solve this....

conservation of energy? define 1 meter up as the base..
then potential energy at the end = mg(21m)
Initial KE = 1/2 mv^2.
the vertical component of KE is sin(35)*v

no, screw that method... you'd have to calculate the vertical component of kinetic energy at the end as well..

Has anyone else ever had a problem this difficult in 11th grade physics??

 

[Wheatmaster]

x=vcos0t
y=vsin0t

 

[JustAnAverageGuy]

Originally posted by: DrPizza
Has anyone else ever had a problem this difficult in 11th grade physics??

I did, but that was last term

 

[minime72706]

Originally posted by: DrPizza
I'll retype it in this reply window..

Anyway, I've come up with 2 equations, 2 unknowns... one of the unknowns is time, the other is the initial velocity.

first, deal with the horizontal part...
Easy equation: d=v*t
Use the horizontal velocity which is a component of the initial velocity...
So, 140 meters = v * cos(35degrees) * t

2nd, deal with the vertical part...
initial velocity is sin(35)*v

to find the time, use d = initial velocity*t + 1/2 a t^2
initial velocity in this case is sin(35)*v

So, you have 21 meters = sin(35)*v + 1/2 (9.81 meters per second squared) * t^2
(leaving off the units...
21 = sin35 *v + .5*9.81*t^2

since you're looking for velocity, substitute for time. Using the first equation, t= 140/(v*cos35) and plug it in for t
Solve for v.

Now im starting to feel dumb, but i dont remember how to solve the the exponent. Wouldn't it be a negitive 9.8 for gravity. The same problem in my book, with different numbers, but it doesn't have the answer of how to solve it. Its Giancoli Physics if that means anything.

 

[User1001]

This is a relatively simple problem:
You separate the problem into components and therefore two equations.
(x component) X=V(init. x velocity)t
(y component) Y=(-1/2)g^2 + V(init Y velocity)t + h(init height)
V(init x)=cos35V(init)
V(init y)=Sin 35 V(init)
V(init x)^2 + V(init y)^2=v(init)^2
That should be enough to work with

 

[TuxDave]

3*10^8 meters/sec. It didn't ask for the minimum initial speed to accomplish it.

 

[TuxDave]

Originally posted by: minime72706

Originally posted by: DrPizza
oh crap... I looked at something, hit the reply button again, and lost what I typed.

Incidentally, if you inspect the problem closely enough, you should realize that there are TWO solutions to the problem...

One for it to have cleared the wall while still on the way up, the other to have cleared the wall on the way down.

its safe to assume that it will land on top if the wall, so only one answer makes sense

ahhh. fine.... make the question a little less creative..

 

[DrPizza]

Originally posted by: minime72706

Originally posted by: DrPizza
I'll retype it in this reply window..

Anyway, I've come up with 2 equations, 2 unknowns... one of the unknowns is time, the other is the initial velocity.

first, deal with the horizontal part...
Easy equation: d=v*t
Use the horizontal velocity which is a component of the initial velocity...
So, 140 meters = v * cos(35degrees) * t

2nd, deal with the vertical part...
initial velocity is sin(35)*v

to find the time, use d = initial velocity*t + 1/2 a t^2
initial velocity in this case is sin(35)*v

So, you have 21 meters = sin(35)*v + 1/2 (9.81 meters per second squared) * t^2
(leaving off the units...
21 = sin35 *v + .5*9.81*t^2

since you're looking for velocity, substitute for time. Using the first equation, t= 140/(v*cos35) and plug it in for t
Solve for v.

Now im starting to feel dumb, but i dont remember how to solve the the exponent. Wouldn't it be a negitive 9.8 for gravity. The same problem in my book, with different numbers, but it doesn't have the answer of how to solve it. Its Giancoli Physics if that means anything.

<reaches over to bookshelf.... ah ha, here it is.. >
Which page?

 

[minime72706]

Originally posted by: DrPizza

Originally posted by: minime72706

Originally posted by: DrPizza
I'll retype it in this reply window..

Anyway, I've come up with 2 equations, 2 unknowns... one of the unknowns is time, the other is the initial velocity.

first, deal with the horizontal part...
Easy equation: d=v*t
Use the horizontal velocity which is a component of the initial velocity...
So, 140 meters = v * cos(35degrees) * t

2nd, deal with the vertical part...
initial velocity is sin(35)*v

to find the time, use d = initial velocity*t + 1/2 a t^2
initial velocity in this case is sin(35)*v

So, you have 21 meters = sin(35)*v + 1/2 (9.81 meters per second squared) * t^2
(leaving off the units...
21 = sin35 *v + .5*9.81*t^2

since you're looking for velocity, substitute for time. Using the first equation, t= 140/(v*cos35) and plug it in for t
Solve for v.

Now im starting to feel dumb, but i dont remember how to solve the the exponent. Wouldn't it be a negitive 9.8 for gravity. The same problem in my book, with different numbers, but it doesn't have the answer of how to solve it. Its Giancoli Physics if that means anything.

<reaches over to bookshelf.... ah ha, here it is.. >
Which page?

I don't know if your serious, but page 70 question 62 in my book(sixth edition), it's chap 3

 

[Martin]

Originally posted by: User1001
This is a relatively simple problem:
You separate the problem into components and therefore two equations.
(x component) X=V(init. x velocity)t
(y component) Y=(-1/2)g^2 + V(init Y velocity)t + h(init height)
V(init x)=cos35V(init)
V(init y)=Sin 35 V(init)
V(init x)^2 + V(init y)^2=v(init)^2
That should be enough to work with

Good answer.

 

[TuxDave]

Originally posted by: DrPizza
I'm trying to think of an easier way to solve this....

conservation of energy? define 1 meter up as the base..
then potential energy at the end = mg(21m)
Initial KE = 1/2 mv^2.
the vertical component of KE is sin(35)*v

no, screw that method... you'd have to calculate the vertical component of kinetic energy at the end as well..

Has anyone else ever had a problem this difficult in 11th grade physics??

I had problems similiar in difficulty in 11th grade physics so it's not too bad. HOWEVER, I don't recall a simplier way to do it besides the way presented in the thread.

 

[DrPizza]

Originally posted by: Martin

Originally posted by: User1001
This is a relatively simple problem:
You separate the problem into components and therefore two equations.
(x component) X=V(init. x velocity)t
(y component) Y=(-1/2)g^2 + V(init Y velocity)t + h(init height)
V(init x)=cos35V(init)
V(init y)=Sin 35 V(init)
V(init x)^2 + V(init y)^2=v(init)^2
That should be enough to work with

Good answer.

what's that (bolded)??
Maybe it's the notation, but that doesn't appear to even make sense.

edit: does he mean Y = (-1/2)g t^2 + V(init Y velocity)t + h ?
And, how is that good enough to work with? It doesn't even address the distance and height of the wall!

 

[TuxDave]

Do we have a final answer yet? I think I have an unorthodox way of solving it and I need to confirm the numbers.

Edit: My method has the solution at 43.105m/s and t=3.965sec w/ g=9.8

 

[minime72706]

I used trial and error on will it clear the fence and got 43.38821256m/s

http://hyperphysics.phy-astr.g...u/hbase/traj.html#tra9

 

[TuxDave]

Originally posted by: minime72706
I used trial and error on this site and got 43.38821256m/shttp://hyperphysics.phy-astr.g...u/hbase/traj.html#tra9">will it clear the fence</a>

Nice... I'll post my method since we have the same answer. It requires no trial and error.

 

[minime72706]

Yes my linking skills do suck

 

[neonerd]

guess mine was wrong

 

[bootymac]

Man, when we did Kinetics in the first term, there wasn't stuff like this! We just started vectors too

But I got the highest mark (105%) on the exam All those nerds that don't know what sex is can suck on my nads

 

[minime72706]

Originally posted by: DrPizza

Originally posted by: Martin

Originally posted by: User1001
This is a relatively simple problem:
You separate the problem into components and therefore two equations.
(x component) X=V(init. x velocity)t
(y component) Y=(-1/2)g^2 + V(init Y velocity)t + h(init height)
V(init x)=cos35V(init)
V(init y)=Sin 35 V(init)
V(init x)^2 + V(init y)^2=v(init)^2
That should be enough to work with

Good answer.

what's that (bolded)??
Maybe it's the notation, but that doesn't appear to even make sense.

edit: does he mean Y = (-1/2)g t^2 + V(init Y velocity)t + h ?
And, how is that good enough to work with? It doesn't even address the distance and height of the wall!

its the fromula y=yo +Vyo*t-.5gt^2
where y is the fina height and yo is the starting height

 

[TuxDave]

I'm basing my solution on this other physics question. Does anyone remember the question that if you're trying to shoot an apple that just starts to fall from a tree, at what angle should you shoot at it. The answered turned out that you want to aim your gun directly at the apple before it starts falling, and as the apple falls, your bullet will curve equally and hit the apple.

Using this same principle, assume you're trying to hit an apple 140m away and some distance off the ground. Your gun is at 35 degrees and when the bullet hits the apple, you want it such that it be 21 meters off the ground. Using the same principle, your gun should be aiming directly at the apple so the initial apple height should be:

140*tan(35) meters

The distance the apple will fall before getting hit 21 meters off the ground is:

140*tan(35)-21=77.029 meters

The time for it takes the apple to fall such a distance is:

0.5*9.8*t^2=77.029 meters
t = 3.965 (using the later of the two solutions).

If you want the bullet traverse 140m horizontally in 3.965secs, your initial velocity (@ 35 degrees) should be:

140m=v*cos(35)*3.965
v = 43.104 m/s

QED

1337fizicks.jpg

 

[DrPizza]

I had a really "simple" way of solving it.. but don't have a trusty calculator with me.

(oh, btw, I have the 5th edition... thought I had the teacher's edition as well, but I don't)

That was to use some simple calculus..
I dropped the problem vertically one point...
then I had 2 points (0,0) and (140,21)

So, starting with y = ax^2 + bx + c
since (0,0), c = 0

dy/dx @ x=0 is tan(35 degrees)
dy/dx = 2ax + b = tan35
b = tan 35

ax^2 + bx + c = y, plug in b, and the other point to find a
a(140^2) + 140tan 35 + 0 = 21
a = (21 - 140tan35)/140^2

and I have the equation for the height at any x.

Then, you can find the x and then y value at the turning point...
since dy/dx = 2ax + b = 0

so x = -tan35 / (2*(21-140tan35)/140^2)

or x = (-140^2 * tan35) / (42-170tan35)

plug this in to solve for y - the maximum height the ball reaches..
find the initial velocity in the y direction
and ultimately, the initial velocity of the ball.

 

[DrPizza]

Originally posted by: minime72706

Originally posted by: DrPizza

Originally posted by: Martin

Originally posted by: User1001
This is a relatively simple problem:
You separate the problem into components and therefore two equations.
(x component) X=V(init. x velocity)t
(y component) Y=(-1/2)g^2 + V(init Y velocity)t + h(init height)
V(init x)=cos35V(init)
V(init y)=Sin 35 V(init)
V(init x)^2 + V(init y)^2=v(init)^2
That should be enough to work with

Good answer.

what's that (bolded)??
Maybe it's the notation, but that doesn't appear to even make sense.

edit: does he mean Y = (-1/2)g t^2 + V(init Y velocity)t + h ?
And, how is that good enough to work with? It doesn't even address the distance and height of the wall!

its the fromula y=yo +Vyo*t-.5gt^2
where y is the fina height and yo is the starting height

that's what I finally figured.. (see what I added) -- he has g^2, not t^2

 

[DrPizza]

Originally posted by: TuxDave
I'm basing my solution on this other physics question. Does anyone remember the question that if you're trying to shoot an apple that just starts to fall from a tree, at what angle should you shoot at it. The answered turned out that you want to aim your gun directly at the apple before it starts falling, and as the apple falls, your bullet will curve equally and hit the apple.

Using this same principle, assume you're trying to hit an apple 140m away and some distance off the ground. Your gun is at 35 degrees and when the bullet hits the apple, you want it such that it be 21 meters off the ground. Using the same principle, your gun should be aiming directly at the apple so the initial apple height should be:

140*tan(35) meters

The distance the apple will fall before getting hit 21 meters off the ground is:

140*tan(35)-21=77.029 meters

The time for it takes the apple to fall such a distance is:

0.5*9.8*t^2=77.029 meters
t = 3.965 (using the later of the two solutions).

If you want the bullet traverse 140m horizontally in 3.965secs, your initial velocity (@ 35 degrees) should be:

140m=v*cos(35)*3.965
v = 43.104 m/s

QED

1337fizicks.jpg

I love that solution! Much simpler than mine....

 

[TuxDave]

Originally posted by: DrPizza
I love that solution! Much simpler than mine....

I loved physics when I was in High School. Too bad I don't get to practice it much in EE.