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Two questions from physics 11 that my teacher cannot figure out. HELP! 
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#1
Mar2212, 04:44 PM

P: 1

1. The problem statement, all variables and given/known data
These are from a british columbia provincial exam from a few years ago. My teacher enlisted my help in trying to figure out a solution. This is supposed to be for physics 11. I can't figure it out. Maybe one of you awesome people can help me. Thanks... Question1. A Quarterback can throw a football a maximum range of 80 meters on level ground. What is the highest point reached by this football if thrown this maximum range? Ignore air friction. a. 30m b. 50m c. 20m d. 10m e. 40m < this is the answer but we cannot figure out why. Question 2. A baseball is thrown vertically into the air at a velocity (v), and reaches maximum height (h). At what height was the baseball moving with one half of its original velocity? air resistance is negligible. a. 0.50 h b. 0.33 h c. 0.25 h d. 0.75 h e. 0.67 h < Correct answer, but how? 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
Mar2212, 05:12 PM

P: 157

Question 1: Use v[itex]^{2}_{2}[/itex]=v[itex]^{2}_{1}[/itex]2ad
You want to find the vertical velocity. You are given the distance it travels. What is the only acceleration acting on the football? 


#3
Mar2212, 07:50 PM

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Neither one of those selected answers appears to be correct. You need to show some attempt at answering the questions.



#4
Mar2212, 08:33 PM

P: 297

Two questions from physics 11 that my teacher cannot figure out. HELP!
Check again if these were stated as answers.
May be your are looking at the wrong set code (if there is one!). Else there appears to be a typo. HINT: both answers stated are wrong (not just one!) And both the correct answers are also a part of options. You need to show us your work for us to help you :) 


#5
Mar2212, 08:42 PM

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I believe that the indicated answer to the first problem is correct.



#6
Mar2212, 08:49 PM

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The second one appears to be wrong though.



#7
Mar2212, 09:23 PM

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For the first one, find the angle the ball must be thrown, to cover the maximum range with a given speed. (This is a well known result, you will probably find it in a textbook if you can't work it out for yourself.).
The second one can be done easily by using kinetic and potential energy. I think both the answers given by the OP are wrong. 


#8
Mar2212, 09:41 PM

P: 963

Maximum range the angle is 45°
[itex]R=u_xt[/itex] [itex]t=\frac {R}{u_x}[/itex] [itex]S_y=u_yt\frac {1}{2}at^2[/itex] [itex]0=uSinθ(\frac {R}{uCosθ})\frac{1} {2}a(\frac {R}{uCosθ})^2[/itex] [itex]u^2=\frac {aR}{2Cos^2θ}[/itex] [itex]v{_y}^2=u{_y}^22as[/itex] [itex]2aS=\frac {aR}{2Cos^2θ}.Sin^2θ[/itex] [itex]S=\frac {aR}{2Cos^2θ}.Sin^2θ.\frac{1}{2a} [/itex] [itex]S=20 [/itex] 


#9
Mar2212, 09:44 PM

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#10
Apr1712, 11:54 PM

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Let's just look at dropping the ball (if you throw it up to a height h with initial velocity v, it will fall back symmetrically. So it's like dropping a ball from height h to attain a velocity v, except now you want to know how far it dropped where the velocity is v/2.)
acceleration is defined as a=(VfVi)/t. Also, the distance something falls is 0.5*g*t^2. Combine the two and you can get V=sqrt(2gh). So let's say that V and h are from the problem. Now you need to make V come out half as big. Well, what do you need to do to make something under a square root come out to 1/2? You need to divide it by 4, that's what. Seems to me that means when the ball has dropped h/4, the velocity will already be v/2, and the answer to #1 should be 0.75h. 0.67 or 2/3 makes no sense at all. http://en.wikipedia.org/wiki/Equatio...a_falling_body http://www.physicsclassroom.com/class/1dkin/u1l5c.cfm #2 came up in AP Physics. I puzzled over it with another teacher, it was tricky and I don't remember exactly. Break the problem in halfLook at only the half where the ball is falling from the top point down to the ground. You know the horizontal velocity is independent from the vertical velocity. It can also be found that yes 45 is the angle for most distance. So actually, this means Vhorizontal=Vverticalinitial, lets call them just v. The free flight time is going to be d=v*t > 80m/2 = v*t > t=40/v. You substitute that time into the equation for something being thrown straight up. Let's try a=(VfVi)/t again. The initial velocity thrown upward would be v, but if we're looking at the downward half then v will be the final vertical velocity when it hits the ground. The initial vertical velocity at the top of the arc is zero. So 9.8=(v0)/t=v/t. If we swap in t=40/v from above, then 9.8=v/(40/v) and I get v=sqrt(9.8*40). That doesn't seem right, but maybe it is. It's certainly the kind of approach I used before. 


#11
Apr1812, 02:35 AM

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For maximum range, the ball is thrown at 45 degrees. That means the vertical component of velocity and horizontal component of velocity are equal. The horizontal component is such that in the time of flight, the ball covers 80m The maximum height is reached in half that time [in the second half it comes down again]. On the way up, it slows to zero velocity, so the average speed is one half the original. with an average speed of half the size, for only half the time interval, the ball will cover 1/4 the distance, so it reaches a height of only 20m  Option C Question 2: The ball starts at some speed, and this reduces to 0 at maximum height in a certain time interval. half way through that time interval, it will be at half the velocity. First half of time interval, average speed is the average of v and v/2 = 3v/4 Second half of the interval, average speed is the average of v/2 and 0 = v/4 The average sped in the first half of the time interval is 3 times the average speed in the second half of the time interval so the distances are in the ratio 3:1, so it is at height 0.75h that the speed is one half the initial speed. so yes, both answers you suggested are wrong. 


#12
Apr1812, 09:19 PM

P: 29

But under acceleration, the vertical velocity is nonconstant, so you can't just take the average.
Jeepers, isn't there a site where someone has plotted this out already? I was Googling but couldn't find one... 


#13
Apr1812, 09:37 PM

P: 29

Aha, tried the magic word "Trajectory"
http://hyperphysics.phyastr.gsu.edu/hbase/traj.html shows a 45° trajectory rises to about 6.8 for a distance of 25, so this would imply about 22m presuming they programmed correctly. AT http://mysite.verizon.net/res148h4j/...jectory2d.html it calculates 28.0142 m/s goes 79.99953125790009m, 19.999882814475022, 4.038538387261198s flight time. (The "averages" get answers like mine, but that is coincidence IMO). 


#14
Apr1812, 09:37 PM

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Under constant acceleration, [itex]\displaystyle v_\text{average}=\frac{v_1+v_2}{2}\,.[/itex] 


#15
Apr1812, 09:57 PM

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#16
Apr1812, 10:03 PM

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EDIT: in the box, lower Right and your answer of 6.8 must be wrong, as the answer will actually be 6.25 if the range is indeed 25m. 6.8 is the approximate answer for a range of 27m In you use you second link and sub in v_{o} = 28.014282 and the angle as 45^{o} you will see the answer. Range = 80, Height = 20 


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