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Physics Help

M

minime72706

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

You have too learned enough physics 😛
 
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...
Click to expand...


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
 
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?
Click to expand...

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


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

Technically between those 2 points are infinitely many solutions, 😉.

Weird problem. Could have been worded better.
 
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.
 
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.
 
you get to use a graphing calculator to solve this? Or, do you have to solve this by hand??
 
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