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Physics Problem - Page 2
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HeavOnEarth
United States7087 Posts
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ketomai
United States2789 Posts
1.82=2.3+5.6154*t+.5*-9.8*t^2 0= -4.9t^2+5.6154t+.48 | ||
micronesia
United States24799 Posts
On September 24 2008 07:19 HeavOnEarth wrote: How'd you know i didn't take a physics class before this? D: Mind reader, you are. I can't tell if you are being sarcastic or not. If yes: meh If no: Kinda obvious from the OP. | ||
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HeavOnEarth
United States7087 Posts
For me it's sarcasm or just friendly joking around :O | ||
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MuShu
United States3223 Posts
You can ignore the weight, since gravity is always -9.8m/s^2 if down is negative. So you have: a=45.7m/s^2 g=-9.8m/s^2 initial v=0 distance to release=.69m converted from cm To find the speed of the shot as it's released, use the third equation posted by nevake, and you'll get: v^2= 0 + 2(45.7)(0.69) + Show Spoiler + v= 7.94 m/s Next I find the time it takes from the beginning to the releasing of the rock. To do that, I use the 1st equation, using the final velocity I found and the acceleration provided 7.94 - 9.8t time to release = + Show Spoiler + .174s To find the time to the highest point, I also use the first equation. I know the final velocity is 0 because at the very top of the flight, the velocity is always zero, since to go from positive to negative velocity, it has to pass through 0 velocity. Also, the acceleration is -9.8m/s^2 since now that the man released the rock, only gravity is acting on it. So you get this equation: 0 = 7.94 - 9.8t time to top = + Show Spoiler + .81s With the time from release to the highest point discovered, we can now use the 2nd equation to find the distance it takes to reach the top distance to top = 0.69 + 7.94(.81) -4.9(.81^2) = + Show Spoiler + 3.91m To find the time it takes to reach the man's head, we again use the 2nd equation, since we have the distance, initial velocity, and acceleration: + Show Spoiler + And there you go. This problem you will consider easy later on after you've practiced how to apply the three equations, and other ones will be pretty much the same, you just need to apply them differently as they add in angles and two directions. edit: I'm solving these algebraically by the way, not with Calculus, since my school only has 3 students who took AP Calc taking AP Physics as seniors, so most other people don't know any Calculus. I looked at some of the other answers ppl posted and they're different than mine, so I dunno, I guess check with your teacher since I don't wanna spend more time on this XD. I did this the most straightforward way I could think of using the 3 equations, and since he was throwing a 16 pound rock at that acceleration for about half a bodylength, 3.91m to the top sounds about right compared to like 10m in one of the other problems. o yea, and make sure to tell us what the real answer is lol | ||
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HeavOnEarth
United States7087 Posts
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SayaSP
Laos5494 Posts
On September 24 2008 07:19 HeavOnEarth wrote: How'd you know i didn't take a physics class before this? D: Mind reader, you are. The units are driving me crazy... everything has to be perfect and u can't use like SI units with... metrics and god damn how are you supposed to remember everything D:::: O well, i'm a glutton for pain ;D He's a certified physics teacher | ||
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Polyphasic
United States841 Posts
On September 24 2008 06:22 HeavOnEarth wrote: Sam heaves a shot with weight 16-lb straight upward, giving it a constant upward acceleration from rest of 45.7 m/s2 for a height 69.0 cm. He releases it at height 2.30 m above the ground. *Meaning the ball released his hand @ 2.30m above the ground* You may ignore air resistance. What is the speed of the shot when he releases it? How high above the ground does it go? How much time does he have to get out of its way before it returns to the height of the top of his head, a distance 1.82 m above the ground? It's not really a homework problem; she just posted it on the board and said try it at home , see if you can solve it ; and i wanna be the cool kid that knows the answer  Of course, I'm not blindly giving you this question without working on it, however i probably messed everything up :D I don't expect anyone to do it but... if you could point me in the general direction that would be pretty awesome [: . ;edit; Excuse my godawful drawing T_T ![[image loading]](http://i38.tinypic.com/16bcwi9.jpg) ;edit2; Also 69cm = (69)1x10^-2 m or .69 m And wow TL is really helpful :DD Also, i suppose that when the ball is coming down ; g ( acceleration), due to gravity = ~9.81m/s^2 Ahh and also this wasn't a homework problem or anything she just gave it to us as a kind of eye opener , she's going to intro./ review it tomorrow in class. So far we've just been doing physics I review so we're starting kinematics, plus the hurricane set us back a week( no school) so that would explain the lack of movement in the class? D: Thanks for your help again ! please try to type the problem correctly so it can be solved. you can't give something a constant acceleration by throwing it. constant acceleration means that it keeps accelerating, which isn't possible if you throw it. if it is as you say, that there is a constant upward acceleration of 45.7m/s2, (and if you take into account that downward acceleration due to gravity is 9.8m/s2), then the result is that the object will continue to accelerate upwards indefinitely. the problem should be straightforward if the 45.7 is velocity rather than acceleration. the object starts by having an upward velocity of 45.7. from then, each second that passes by, it loses 9.8 of its original velocity due to gravity. this means that it goes upwards for (45.7/9.8) seconds before it reaches its highest point. => so if while going upwards, its time spent is 45.7/9,8, and its average velocity is 45.7/2, you can figure out the distance. => now you can find the total height at the highest point. from there, you can find how long it takes for the object to reach a height of 1.82. | ||
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Hippopotamus
1914 Posts
On September 24 2008 11:19 Polyphasic wrote: please try to type the problem correctly so it can be solved. you can't give something a constant acceleration by throwing it. constant acceleration means that it keeps accelerating, which isn't possible if you throw it. if it is as you say, that there is a constant upward acceleration of 45.7m/s2, (and if you take into account that downward acceleration due to gravity is 9.8m/s2), then the result is that the object will continue to accelerate upwards indefinitely. the problem should be straightforward if the 45.7 is velocity rather than acceleration. the object starts by having an upward velocity of 45.7. from then, each second that passes by, it loses 9.8 of its original velocity due to gravity. this means that it goes upwards for (45.7/9.8) seconds before it reaches its highest point. => so if while going upwards, its time spent is 45.7/9,8, and its average velocity is 45.7/2, you can figure out the distance. => now you can find the total height at the highest point. from there, you can find how long it takes for the object to reach a height of 1.82. You say all this in light of a posted solution on the 1st page? | ||
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HeavOnEarth
United States7087 Posts
please try to type the problem correctly so it can be solved. but but i typed out the 69cm , it only goes up to 69 cm if it is as you say, that there is a constant upward acceleration of 45.7m/s2, (and if you take into account that downward acceleration due to gravity is 9.8m/s2), then the result is that the object will continue to accelerate upwards indefinitely. noooh it stops after the .69m thing im sry ill make it clearer even though i bolded it and drew a picture and everything;edit; im soo confused now T.T | ||
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Polyphasic
United States841 Posts
if the ball accelerates for .69 meters, after which its acceleration is 0 (presumably because it took .69 meters to throw the ball). 1- find the time it takes for the ball to go .69 meters if it starts at rest, and its acceleration is 45.7m/s2. (this is an equation you use) 2- after that, the scenario is that the ball starts from the end of the throw (starting point plus .69 meters) at the velocity you calculated earlier. from this point, the solution is same as my previous post. =) don't take my toneless post too hard. i tend to be very friendly in real life. | ||
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HeavOnEarth
United States7087 Posts
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micronesia
United States24799 Posts
Only conditionally initially: http://www.highered.nysed.gov/tcert/certificate/typesofcerts.htm#ci | ||
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