<p>Maybe it's because I'm over-analyzing it but I've wasted 20 sheets of paper trying to get the right answer for this problem, but I just dont get it!!!!!</p>
<p>A shot put is released with a velocity of 12 m/s and stays in the air for 2.0s.</p>
<p>At what angle with the horizontal was it released?</p>
<p>What horizontal distance did it travel?</p>
<p>if it stays in the air for two secs,</p>
<p>vy=voy+ayt; ay=-9.8m/s^2 and t=1s (for going up)
0=voy+(-9.8m/s^2)(1s)
voy=9.8m/s</p>
<p>v=(vx^2+vy^2)^.5
12m/s= (vx^2+(9.8m/s)^2)^.5
solve for vx</p>
<p>angle= tan-1(vy/vx)</p>
<p>horizontal distance
delta x= vxt (acceleration in the x direction is 0)</p>
<p>and solve. I don’t have a calculator on me</p>
<p>ok so what’s the x value? In my class we only use v initial and v final, so I’m a bit confused, lol, sorry about that.</p>
<p>"A shot put is released with a velocity of 12 m/s and stays in the air for 2.0s.</p>
<p>At what angle with the horizontal was it released?</p>
<p>What horizontal distance did it travel?"</p>
<p>For the vertical component, </p>
<p>v<em>initial = 12 sin x
v</em>final = 0
a = -9.8 m/s^2
t = 1s</p>
<p>Vf = Vi + at
0 = 12 sin x + -9.8(1)
-12 sin x = -9.8
12 sin x = 9.8
x = 54.75°</p>
<p><< Angle to horizontal = 54.75° >></p>
<p>Horizontal distance = V<em>x</em>component * t
d = 12 cos 54.75 * 2
<< d = 13.85 m >></p>
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<p>Wow! Thanks aisdufhb! Thanks for your lucid explanation Bassir, And College-goer, thank you so much, that really made sense, becasue those are the variables that my class uses. Now I’m sure my Physics skills with get a 100:) Thanks again guys!</p>
<p>Dude, no offense, but if that problem gave you so much trouble, the AP’s gonna be a bloodbath…</p>
<p>Yeah, I think it was really easy.</p>
<p>A flaw in the problem is that it is assumes that an ant or other nearly-zero-height creature is throwing the shot-put :)</p>
<p>it could be shot out of the ground.</p>
<p>btw,</p>
<p>vo (Velocity @ 0)= initial velocity
vx = velocity in the x-direction</p>