Posted: 5/12/2009 7:29:25 PM EDT
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She's studying for her final, and the prof gave her a study guide. She's stuck on one of the questions, and I told her that ARFCOM could solve it within 10 minutes.
Again, this is a study guide, so she isn't cheating. Question: During a tennis match, a player servers the ball at a speed of 32m/s completely in the horizontal direction from a height of 1.8m above the ground. The net is located 12m away. What is the angle at which the ball is traveling at the moment it reaches the net (ball should be 1.1m above the ground at this point, she says)? Give your answer in degrees measured down from the horizontal. The redhead awaits your response. |
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Assuming there is no real drag effect, the ball traveling at 32 m/s takes 12/32 seconds to reach the net. It is also free falling from 1.8 meters above the ground under 9.8 meter/second2. So at the net, it is moving 32 m/s in the X direction and 9.8*12/32 m/s in the Z (vertical axis)....that is 3.68 m/s (significant digits). So the angle is arctan (3.68/32) or 6.55 degrees. The drop of the ball is 9.8*1/2(12/32)2 or 0.69 meters, again with significant digits.... |
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think outside the bun you have a triangle already with the measurements there. you have the distance away from the net the start height and end height, there is a 90deg angle in that triangle. the deg should be found by completing the triangle. I was thinking this as well. The velocity seems like unnecessary information to me. |
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think outside the bun you have a triangle already with the measurements there. you have the distance away from the net the start height and end height, there is a 90deg angle in that triangle. the deg should be found by completing the triangle. sounds good to me. just basic trig |
| I believe it would be a vector problem. I forget the formula used to find the speed of a falling object at 1g in a certain period of time, but I do know that things accelerate at 9.8m/s/s. So find out how long it takes something going 32 m/s to go 12 meters and plug that into whatever formula it is, and then use that vertical velocity and the 32m/s/s horizontal velocity to create a right angle vector. Then just measure the hypotenuse with the pythagorean therom. I don't know enough about those things to figure it out right now, I just know how it would be done. |
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Posted: Today 11:50:31 PM EDT by mcg3686 think outside the bun you have a triangle already with the measurements there. you have the distance away from the net the start height and end height, there is a 90deg angle in that triangle. the deg should be found by completing the triangle. +1 velocity in x-direction = v_x (t) =constant v_y(t) = 1/2 * g * t^2, where g is the acceleration of gravity angle (t) = arctan[v_y(t)/v_x(t)] Maybe it's a trick question. What's the height of a tennis net? 
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Assuming there is no real drag effect, the ball traveling at 32 m/s takes 12/32 seconds to reach the net. It is also free falling from 1.8 meters above the ground under 9.8 meter/second2. So at the net, it is moving 32 m/s in the X direction and 9.8*12/32 m/s in the Z (vertical axis)....that is 3.68 m/s (significant digits). So the angle is arctan (3.68/32) or 6.55 degrees. The drop of the ball is 9.8*1/2(12/32)2 or 0.69 meters, again with significant digits.... Thanks, she says that makes sense and is one of the possible answers. It's been a while since I did that kind of physics. |
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My thought processing can't begin to work until I see a picture of said girlfriend. BOTD Redhead post. She's there. And thanks, mcg3686. I relayed that to her. what page? Yea, this, couldn't find it. Also, does the carpet match the drapes? |
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Assuming there is no real drag effect, the ball traveling at 32 m/s takes 12/32 seconds to reach the net. It is also free falling from 1.8 meters above the ground under 9.8 meter/second2. So at the net, it is moving 32 m/s in the X direction and 9.8*12/32 m/s in the Z (vertical axis)....that is 3.68 m/s (significant digits). So the angle is arctan (3.68/32) or 6.55 degrees. The drop of the ball is 9.8*1/2(12/32)2 or 0.69 meters, again with significant digits.... No, things ACCELERATE at 9.8m/s/s. It would not travel 12/32 of 9.8 meters in 12/32 of a second. It is not linear, it is exponential (9.8m/s^2). If it were falling linearally, as in 9.8m/s, then you'd be right. An object will actually fall less than 9.8 meters in one second, but at the 1 second mark it will be falling at 9.8m/s. |
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Quoted: Quoted: Assuming there is no real drag effect, the ball traveling at 32 m/s takes 12/32 seconds to reach the net. It is also free falling from 1.8 meters above the ground under 9.8 meter/second2. So at the net, it is moving 32 m/s in the X direction and 9.8*12/32 m/s in the Z (vertical axis)....that is 3.68 m/s (significant digits). So the angle is arctan (3.68/32) or 6.55 degrees. The drop of the ball is 9.8*1/2(12/32)2 or 0.69 meters, again with significant digits.... Thanks, she says that makes sense and is one of the possible answers. It's been a while since I did that kind of physics. It is right. Although I passed Tech Physics I 17 years ago, I use those equations daily. See the last line of my sig....yeah, I got an ass because a few people have ignored me, only to have Murphy bite them in the ass. Then they blamed it on my work but I keep documentation as I have learned...  |
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She has a horizontal velocity (which is a constant velocity, neglecting air resistance).
She has a vertical velocity, which starts at zero and accelerates at -9.81 m/s^2. This is a constant acceleration problem. Considering just the horizontal velocity, she can find the time it takes to reach the net. t = d / v Using that time, she can find the vertical (downward) velocity at that time. v = v(0) + a * t, and since V(0) = 0, v = a * t The constant horizontal velocity and the vertical velocity are two components of a complete velocity vector. It is trig at that point. Each component (horiz. and vertical velocity) is one side of a right triangle, the hypotenuse is the complete velocity vector. Set it up graphically, decide what trig function to use to find the angle down from the horizontal. ETA: if I didn't get my back of the envelope calculation wrong, she should find the angle to be 6.56° below the horizontal. |
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Quoted: Quoted: Assuming there is no real drag effect, the ball traveling at 32 m/s takes 12/32 seconds to reach the net. It is also free falling from 1.8 meters above the ground under 9.8 meter/second2. So at the net, it is moving 32 m/s in the X direction and 9.8*12/32 m/s in the Z (vertical axis)....that is 3.68 m/s (significant digits). So the angle is arctan (3.68/32) or 6.55 degrees. The drop of the ball is 9.8*1/2(12/32)2 or 0.69 meters, again with significant digits.... No, things ACCELERATE at 9.8m/s/s. It would not travel 12/32 of 9.8 meters in 12/32 of a second. It is not linear, it is exponential. An object will actually fall less than 9.8 meters in one second, but at the 1 second mark it will be falling at 9.8m/s. Assuming the ball is moving 0 m/s in the Z axis when struck, it is moving at 9.8*12/32 m/s after traveling to the net. Acceleration multiplied by time is VELOCITY. But distance, yes, you are correct. That is one-half acceleration multiplied by time SQUARED. The one-half is the INTEGRATION FACTOR from the Calculus. I KNEW this was non-Calc. physics. I know a few physics instructors as there are a few in the family so I learned it well. |
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Quoted: 1st find the time it takes for the ball to reach the net. Using that time find how much the ball falls. now you take arctan(distance ball falls/distance horiz) now you have degrees from the vertical in which that ball hit the net. That is the AVERAGE trajectory. The question was INSTANT trajectory which is a function of the component velocities. |
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Assuming there is no real drag effect, the ball traveling at 32 m/s takes 12/32 seconds to reach the net. It is also free falling from 1.8 meters above the ground under 9.8 meter/second2. So at the net, it is moving 32 m/s in the X direction and 9.8*12/32 m/s in the Z (vertical axis)....that is 3.68 m/s (significant digits). So the angle is arctan (3.68/32) or 6.55 degrees. The drop of the ball is 9.8*1/2(12/32)2 or 0.69 meters, again with significant digits.... No, things ACCELERATE at 9.8m/s/s. It would not travel 12/32 of 9.8 meters in 12/32 of a second. It is not linear, it is exponential. An object will actually fall less than 9.8 meters in one second, but at the 1 second mark it will be falling at 9.8m/s. Assuming the ball is moving 0 m/s in the Z axis when struck, it is moving at 9.8*12/32 m/s after traveling to the net. Acceleration multiplied by time is VELOCITY. But distance, yes, you are correct. That is one-half acceleration multiplied by time SQUARED. The one-half is the INTEGRATION FACTOR from the Calculus. I KNEW this was non-Calc. physics. I know a few physics instructors as there are a few in the family so I learned it well. Oh, I read the post wrong. Yeah, that's the velocity. I was reading it as that being how to find out how FAR it fell. I was confused in my other post too. I was mixing the idea of distance vs. velocity and confusing myself. I was right, but I was thinking that you need some formula to figure out the VELOCITY, when velocity is linear. Oh well, I've had a few beers. 
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My thought processing can't begin to work until I see a picture of said girlfriend. BOTD Redhead post. She's there. And thanks, mcg3686. I relayed that to her. what page? Yea, this, couldn't find it. Also, does the carpet match the drapes? Pages 10 and 14, and I already said she was natural. But...what carpet? 
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Quoted: Quoted: Quoted: Wow, you're dating a high school sophomore. Wow, you have no class. OK, maybe just a bit  Unless she is in a private university, this sounds like a junior college as all major state universities have already taken finals. . I know, my bro had a bitch fest on the phone tonight as one of his students missed the final and now wanted a makeup...Don't be so harsh. Some people are non-traditional students. And they are typically the better students. |
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Wow, you're dating a high school sophomore. Wow, you have no class. OK, maybe just a bit Unless she is in a private university, this sounds like a junior college as all major state universities have already taken finals. . I know, my bro had a bitch fest on the phone tonight as one of his students missed the final and now wanted a makeup...
Don't be so harsh. Some people are non-traditional students. And they are typically the better students. Actually, it's a small state school. Finals are all this week. I just finished picking up a CS IA minor (well, not officially, I didn't ever have a major there, so they won't add on a minor to my degree) at the same school. |
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My thought processing can't begin to work until I see a picture of said girlfriend. BOTD Redhead post. She's there. And thanks, mcg3686. I relayed that to her. what page? Yea, this, couldn't find it. Also, does the carpet match the drapes? Pages 10 and 14, and I already said she was natural. But...what carpet? haha, great answer. At this point, I would begin to help, but the question has been answered. |
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Wow, you're dating a high school sophomore. Wow, you have no class. OK, maybe just a bit Unless she is in a private university, this sounds like a junior college as all major state universities have already taken finals. . I know, my bro had a bitch fest on the phone tonight as one of his students missed the final and now wanted a makeup...
Don't be so harsh. Some people are non-traditional students. And they are typically the better students. Dammit, I did the equation out all nice and stuff and shoulda guessed Keith J beat me to it!
And actually Keith, the University of Iowa is in its finals week - my first is tomorrow. Graduation is Saturday 
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Quoted: She's studying for her final, and the prof gave her a study guide. She's stuck on one of the questions, and I told her that ARFCOM could solve it within 10 minutes. Again, this is a study guide, so she isn't cheating. Question: During a tennis match, a player servers the ball at a speed of 32m/s completely in the horizontal direction from a height of 1.8m above the ground. The net is located 12m away. What is the angle at which the ball is traveling at the moment it reaches the net (ball should be 1.1m above the ground at this point, she says)? Give your answer in degrees measured down from the horizontal. The redhead awaits your response. SIIHPAPP |
