Physics Help

[minime72706]

A base ball clears a fence 140m away and 22m tall and was hit at angle of 35degrees what is the intial speed?
Assume it was hit from a height of 1m

I dont need an answer, just an explination how to find it, like what formulas to use.
Thanks

 

[FreshFish]

Your book should have some equations for motion under a constant acceleration (gravity); those should be able to provide the answer, given that you break it into horizontal and vertical components.

 

[minime72706]

My book does have formulas, but none with two unkowns, time and velocity.

 

[User1001]

kinematics

 

[JustAnAverageGuy]

You have to use two formulas.

One to find Y, then plug Y into the one to find X.

I'll see if I can find my notes

 

[neonerd]

Originally posted by: JustAnAverageGuy
You have to use two formulas.

One to find Y, then plug Y into the one to find X.

I'll see if I can find my notes

ditto

 

[maziwanka]

i think your missing some info

 

[neonerd]

x and y are independant in this one:

to find y use: Vfy^2 = Viy^2 - 2g(delta)y

to find the x component, use: Vx = (delta)x/(delta)t

i think you may me missing some info as said above by maziwanka

 

[bootymac]

I tried to figure it out, but I havn't learned enough to help you (Grade 11 Physics only)

 

[JustAnAverageGuy]

Image

So it looks like this, correct?

 

[DrPizza]

do you mean... "just cleared"? in terms of making it over the fence?

 

[minime72706]

Originally posted by: JustAnAverageGuy
Image

So it looks like this, correct?

Yes, but its is a arc, not a straight line

 

[DrPizza]

I'm going to assume you mean just cleared... There are probably a multiple number of ways to solve this problem

more info needed though

Also, is this calculus based or formula based?


btw, is this an extra credit assignment?? Seems like a more difficult problem than I'd expect...

 

[DrPizza]

Originally posted by: bootymac
I tried to figure it out, but I havn't learned enough to help you (Grade 11 Physics only)

You have too learned enough physics

 

[minime72706]

Originally posted by: DrPizza
I'm going to assume you mean just cleared... There are probably a multiple number of ways to solve this problem

more info needed though

Also, is this calculus based or formula based?


btw, is this an extra credit assignment?? Seems like a more difficult problem than I'd expect...

This is formula based, i'm in grade eleven physics so no calc yet
Its just a normal assingment
are teacher did something like it, and used found two times and set them equal, but i cant figure it out

 

[JustAnAverageGuy]

X= V[x0]T

Y = V[y0]T + (-9.8)(T²) / 2

Familiar or am I thinking of the wrong kind of problem?

 

[minime72706]

Originally posted by: JustAnAverageGuy
X= V[x0]T

Y = V[y0]T + (-9.8)(T²) / 2

Familiar or am I thinking of the wrong kind of problem?

yes, but basiclly it lands on top of the wall, so i need to figure in height change

 

[FriedPixel]

hmm

 

[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.

 

[Hayabusa Rider]

Gah didnt read the problem correctly

 

[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

 

[BigPoppa]

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.

Technically between those 2 points are infinitely many solutions, .

Weird problem. Could have been worded better.

 

[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.

 

[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.

edit: incidentally, in the post above, d=initial velocity *t + 1/2 a t^2
d is 21 because it goes up from 1 m to 22 meters.

 

[DrPizza]

you get to use a graphing calculator to solve this? Or, do you have to solve this by hand??