<p>I'm reviewing for midterms and can't seem to crack a few problems. The two worst of these are the following:</p>
<p>Projectile</a> Motion: Is it me or is the time variable insufficient data to solve? I'm retarded...</p>
<p>Impulse:</a> I thought the answer was 144. Ft = ΔP = (18)(8) = 144 N·s</p>
<p>Any help would be appreciated. And sorry for dragging down the bell curve of intelligence here at CC. :(</p>
<p>Oh yeah AP Physics midterms are coming up... too bad I haven't done like any work on them winter break cause of college stuff. Ughhh... I'm going to have to go back and go through projectile motion and forces and impulse and all that stuff? Great.</p>
<p>For the projectile motion question: You would use X = Vo + at. So X = 9.8(4.5).</p>
<p>well if there was no acceleration at all
like in space
the distance would be zero
so the angle doesnt matter
all that matters is its being pulled away at 9.8m's^2
for 4.5 seconds and u need to find out how far it goes in 4.5 seconds
considering it starts from rest from the line</p>
<p>I would help you with impulse, but I haven't studied it.</p>
<p>Ahh I have to review this stuff, my midterm for physics is on Friday. hehehe I did nothing over break for review.</p>
<p>uhh for the impulse one, isn't it just the area under the graph?</p>
<p>so it would be 20 N * s</p>
<p>for part 2</p>
<p>you find the average force exerted during those 8 seconds which = impulse/time so 20/8 = 2.5 N</p>
<p>the only force is the applied force so F = ma
2.5 N = 2.7 kg * a
a = .926 m/s^2 in the x direction</p>
<p>v = vo + at
v = .926 m/s^2 * 8s = +7.4 m/s</p>
<p>7.4 + (-5.3) = 2.1</p>
<hr>
<p>hope that's right. I haven't done impulse in a while. thanks for posting this, it's a reminder I need to start physics hw.</p>
<p>btw how come you're never on AIM anymore Pater?</p>
<p>Thanks, athlonmj. Always there with clutch physics solutions when I need 'em. :)</p>
<p>michael_pham, I don't quite get what you were saying. Anyway, I'm still struggling with the projectile motion problem.</p>
<p>For the projectile one:</p>
<p>Let v be the vertical component of the velocity with which the projectile is fired. I am not using the entire velocity because the horizontal component is irrevelent. </p>
<p>If there was no gravity, then the initial velocity would keep the projective moving in an upward direction. The distance the projectile went up would be calculuated by multiplying velocity by time (x=vt). </p>
<p>Since gravity is accelerating the projectile downward, the equation for x changes to (x=vt+.5at^2). You need to find the difference between these two x equations at 4.5 sec, so subtract the equations to get /X = -.5at^2. Plug in a and t to find /X</p>
<p>Awesome, tanman. What was stumping me was the basic method for finding the straight-line height--Vyo t.</p>
<p>Thanks a Million,
Brine</p>
<p>well i was bored so i thought id do them for fun. i did part 2 on the impulse a bit different and it was really straightforward. find starting momentum (negative) and then add the impulse (area under the curve) to find the new momentum. divide the new momentum by the mass to get final velocity. pretty simple</p>
