Thank you all for your help. From the answer given here I guessed correctly.
This should have been a really easy problem but the thing was that I missed
4 days of notes and lectures because I was out of town touring a college in
Georgia. So not being at school hurt me but I remembered a lot from last
year. I found the verticle and horizontal components of the velocity and the
time to the fence and even how far the ball was going to travel overall.
Trying to remember what I did I believe I messed up when trying to find the
maximum height reached. From what I can remember I had a number in the
200's. I might have just missed a decimal point somewhere.
One quick question: You used the equation: s = ut + 1/2 at^2... we have
something similar to this in our book but with something different than S
and U. Can you tell me what they stand for.
Thanks again, Steve
oh yeah, how hard is physics in college compared to this kind of stuff you
learn in high school? What kind of careers can you get into with a physics
degree without teaching or just researching stuff?
"Dave Baker" <pumaracing@aol.comNoEmails> wrote in message
news:20041006061132.23043.00001712@mb-m12.aol.com...
Subject: Can someone show me how to answer this problem...
From:
(Steve B.)
Date: 06/10/04 01:40 GMT Daylight Time
Message-id: <cfd9fb63.0410051640.43fe1d8d@posting.google.com>
I'm a senior in high school and the following question was on our test
today for AP Physics 2. I don't know what I was missing but I spent
over 40 minutes on it and still didn't come up with an answer. Can
somehow tell me what the answer is and how to get there?
THE PROBLEM:
A baseball leaves a bat with a speed of 44 m/s and at an angle of 30
degrees above the horizontal. A 5m high fence is located at a
horizontal distance of 132m from the point where the ball is struck.
Assuming the ball leaves the bat 1m above ground level, by how much
does the ball clear the fence?
If someone can help me I'd appreciate it!!
Thank you,
Steve
Firstly always state your assumptions to the examiner. We'll ignore wind
resistance and treat g as being 9.81 m/s^2. It may be that you are told to
use
a different value for g such as 10m/s^2. Make sure you don't miss any such
requirements, especially if they are printed at the start of the question
paper. For extra marks you might want to note to him that in a civilized
country this would be a question about a cricket ball and bat not baseball
ones
:)
Then split the ball's speed into its horizontal and vertical components.
Vh = 44 cos 30 = 38.1051m/s
Vv = 44 sin 30 = 22 m/s
Flight time to the fence is therefore 132/38.1051 = 3.4641 seconds
Now we need to look at the vertical motion of the ball during this time.
We can go a couple of ways here, the short way or the long one. We'll do
both
as a double check on each other and to show how the various Newtonian
equations
can be used.
The ball starts upwards at 22m/s. How high does it travel and how long
does
that take?
Using the equation v = u + at
0 = 22 - 9.81t
t = 2.2426 seconds. The ball rises for 2.2426 seconds.
Using the equation s = ut + 1/2 at^2
It rises for 1/2 x 9.81 x 2.2426^2 metres = 24.6685m. Add the 1m it
started
above the ground and it reaches apogee at 25.6685m
It now starts to fall again for 3.4641 - 2.2426 = 1.2215 seconds
In this time it falls
1/2 x 9.81 x 1.2215^2 = 7.3186m
It's final height as it reaches the fence is therefore 25.6685 - 7.3186 =
18.35m. It clears the fence by 13.35m
However, the equation s = ut + 1/2 at^2 can be used to describe the entire
trajectory of the ball, not just its upwards or downwards motion
separately.
But we need to get the signs of the speed and acceleration correct.
s = (22 x 3.4641) - (1/2 x 9.81 x 3.4641^2)
s = 76.2102 - 58.8599 = 17.35m
Add the 1m it started above the ground = 18.35m. It clears the fence by
13.35m
Both methods agree. Phew. You're on your way to university :)
--
Dave Baker - Puma Race Engines (www.pumaracing.co.uk)
.
