Mon-Thurs Block 10/28-31 Projectile Motion
Post date: Oct 28, 2013 8:55:29 PM
We got the Projectile motion notes to read over for Tuesday.
We corrected our Formal Set on Simplifying radicals on Tuesday and need to watch the projectile motion video at the bottom of this page in preparation for block class.
Block periods we did the notes on projectile motion below and took our quiz skills 11, 13, 14, & 16 second half of class.
Notes:
A "projectile" is any object that is thrown, shot, or dropped. Usually the object is moving straight up or straight down.
In general, the equation for projectile motion is : h(t) = –16t2 + v0t + h0
"v0" is the initial velocity, "h0" is the initial height, "16" is derived from the value of the force of gravity, and "t" is the time in seconds.
with gravity, the projectile falls 16t2 feet in t seconds.
without gravity to pull the projectile down, its height h would increase according to the equation h(t) = v0t + h0
Examples:
1- An object in launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. What will be the object's maximum height? When will it attain this height?
They didn't give me the equation this time. But that's okay, because I can create the equation from the information that they did give me. The initial height is 80 feet above ground and the initial speed is 64 ft/s. So my equation is:
h= –16t2 + 64t + 80
They want me to find the maximum height. For a negative quadratic like this, the maximum will be at the vertex of the upside-down parabola. So they really want me to find the vertex. From graphing, I know how to find the vertex; in this case, the vertex is at (2, 144):
h = –b/2a = –(64)/2(–16) = –64/–32 = 2
k = s(2) = –16(2)2 + 64(2) + 80 = –16(4) + 128 + 80 = 208 – 64 = 144
But what does this vertex tell me? According to my equation, I'm plugging in time values and extracting height values, so the input "2" must be the time and the output "144" must be the height.Copyright © Elizabeth Stapel 2004-2011 All Rights Reserved
It takes two seconds to reach the maximum height of 144 feet.
2. After the semester is over, you discover that the math department has changed textbooks so the bookstore won't buy back your nearly-new book. You and your friend Herman decide to get creative. You go to the roof of a twelve-story building and look over the edge to the reflecting pool 160 feet below. You drop your book over the edge at the same instant that Herman chucks his book straight down at 48 feet per second. By how many seconds does his book beat yours into the water?
Our initial launch heights will be the same: we're both launching from 160 feet above ground. And the gravity number will be 16. My initial velocity is zero, since I just dropped my book, but my buddy Herman's velocity is a negative 48, the negative coming from the fact that he chucked his book down rather than up. So our "height" equations are:
mine: s(t) = –16t2 + 160
his: s(t) = –16t2 – 48t + 160
In each case, I need to find the time for the books to reach a height of zero ("zero" being "ground level"), so:
mine: 0 = –16t2 + 160, t2 – 10 = 0, so t = ± sqrt(10)
his: 0 = –16t2 – 48t + 160, t2 + 3t – 10 = 0, (t + 5)(t – 2) = 0, so t = –5 or t = 2
I will ignore the negative time values. His book hits the water after two seconds, and mine hits after sqrt(10) seconds, or after about 3.16 seconds. That is:
Herman's book hits the water about 1.16 seconds sooner than mine does.
HW: Projectile Motion WS due Monday, 11/4
Other resources: