<p>It's a pretty cool school, and I'd love to get in! The test tests basic knowledge, but I found this question really difficult. I don't remember ever hearing about such a phenomenon LOL.
Although I'm pretty sure engineering/ prospective engineering majors will find his rudimentary</p>
<p>Anyway...</p>
<p>Also, any crash course links you would recommend? </p>
<p>Look up the units for electric field… :)</p>
<p>…huh?</p>
<p>@noimagination </p>
<p>Its physics , more specifically electricity and magnetism, and and to be more specific voltage</p>
<p>google “the relationship between electric fields and voltage” electric potential is a fancy name for voltage</p>
<p>1j/c (joule per columb) is equal to 1 volt</p>
<p>that’s more than enough direction for to start looking for how to solve, and if in your studies you find something you don’t understand go and google that, do a top down approach to the material (electricity and magnetism) if you have to</p>
<p>Alright thanks…</p>
<p>Now can someone solve it so that i can check my work?</p>
<p>Also, can I ask any of you about a stupid physics question or is it considered taboo? :P</p>
<p>(I am reviewing my SAT physics for the test)</p>
<p>In an SAT practice book, when the book wanted to explain a problem regarding freefall, it was something like this:</p>
<p>Let’s look at a problem where an object is thrown into the air and
returns.
A girl throws a baseball straight up and catches the ball
at the same height 2.6 seconds later. How fast did the girl
throw the ball into the air?</p>
<p>In the solution, the book assumed Vf to be 0 (which I wholly agree with) and used the formula "s = vi t + 0.5gt^2 "</p>
<p>When the book wanted to discuss a problem concerning CURVILINEAR MOTION it was something like this:</p>
<p>Consider a fisherman making a long cast. His lure is released as the
rod moves forward on the cast. The lure rises into the air while at the
same time moving forward across the water. The lure then falls to the
water as it reaches the fisherman’s target.
After careful observation, one can state the following information
about the fisherman’s cast (above): The lure always leaves the rod
tip (considered ground level) at an angle of 30° and a velocity of 22
m/s. How far does the fisherman cast his lure?</p>
<p>In the solution however, when they wanted to exclude Vy (which was calculated to be 11 m/s) and treat it as a freefall problem in its own to find the time taken, they did not take Vyfinal as “0” but rather as 11 m/s. Vyinitial as -11 m/s</p>
<p>and they found the time by (Vyfinal - Vyinitial)/ acceleration (linear motion equation rearranged)</p>
<p>Anyway so my question is, why did they take vyfinal as a value not as 0?! Is it a mistake? or is it? </p>
<p>Is Emags even taught in highschool? </p>
<p>Sure you aren’t lying to us and crunching for a summer class buddy!?</p>
<p>Listen -_- why the heck would I lie to you…</p>
<p>The school’s name is Zewail City (You know Noble prize winning zewail?!)</p>
<p>Here is the electronic copy of their sample test. Not MY problem they included things you say wasn’t taught in high school </p>
<p><a href=“https://www.zewailcity.edu.eg/website/wp-content/uploads/2013/12/Admission-Examination-Sample-English.pdf”>https://www.zewailcity.edu.eg/website/wp-content/uploads/2013/12/Admission-Examination-Sample-English.pdf</a>
@ImUrHuckleBerry </p>
<p>
How about you post your work, and we will critique?</p>
<p>
It was not a mistake - the equations for kinematics they used ignore the throw and catch, they essentially just calculate a parabolic given a single point on that arc. And at the instant it reaches the “end” point it is moving at 11m/s (in their coordinate system).</p>
<p>
It was in mine!</p>
<p>@cosmicfish so why not assume that the “end” point in the first problem is a certain speed as well?! The “end” point was taken as 0 in the first problem. </p>
<p>my point is when do I ignore and when do I not ignore the throw and catch?! It makes a HUGE difference! and both problems are essentially the same!</p>
<p>As long as you clearly define your setup (correctly) such as coordinate systems, vector values, then you will arrive at the correct answer. If you need practice with projectile motion problems, there are a plethora of resources that will more eloquently and thoroughly explain the problem setup better than what one of us could write. </p>
<p>Khan Academy –> <a href=“Physics library | Science | Khan Academy”>https://www.khanacademy.org/science/physics</a></p>
<p>You are ignoring the throw and catch in BOTH problems! I believe you are getting caught up in a notation issue. Let me show you using consistent notation and frames of reference:</p>
<p>1) In this problem, there is only vertical motion - nothing horizontal. Let’s call Z the vertical position (0 at the point of throw/catch), Vz the vertical velocity, and Az the vertical acceleration, all of them positive in the direction of the throw. Note that X, Vx, and Ax (the horizontal components) all equal zero.</p>
<p>At t = 0:</p>
<p>Z = Z0 = 0
Vz = Vz0 = our answer
Az = constant = -9.8 m/s^2</p>
<p>At t = 2.6s</p>
<p>Z = Z1 = 0
Vz = Vz1 = we don’t care!</p>
<p>So using Z1 = Z0 + Vz0<em>t + 0.5</em>Az<em>t^2, you have an equation with only one unknown, which allows you to solve easily for Vz0 = -0.5</em>(-9.8)*2.6^2 = 33.1 m/s</p>
<p>2) We have both horizontal AND vertical motion. Here V = 22m/s at 30 deg above the ground. So:</p>
<p>At t = 0:</p>
<p>X = X0 = 0
Z = X0 = 0
Vx = Vx0 = 22<em>cos(30) = 19.1 m/s
Vz = Vz0 = 22</em>sin(30) = 11 m/s
Ax = constant = 0
Az = constant = -9.8 m/s^2</p>
<p>We are interested at the point where Z1 = Z0, and so we are again solving Z1 = Z0 + Vz0<em>t + 0.5</em>Az<em>t^2 only this time we know V0 and need to find t. We find t = -Vz0 / (0.5</em>Az) = 2.245s.</p>
<p>That described the vertical motion, the horizontal motion is quite simple. Using the same equation with horizontal variables, we have X1 = X0 + Vx0<em>t + 0.</em>Ax<em>t^2, which just becomes X1 = Vx0</em>t = 19.1 * 2.245 = 42.8m.</p>
<p>Note that the acceleration / deceleration due to the catch/throw do not appear in either solution - we care about the ball only when it is solely accelerated by gravity! Does that clarify anything?</p>
<p>I cant imagine the emags classes are very in depth then … Are they really teaching multivariable calc in high school? Because I can’t imagine being able to really derive too much witout gradients and all those special integral theorems you learn at the end of calc III</p>
<p>No, it is all simplified, and in many ways that makes it much, much harder. Didn’t really understand the material til college.</p>
