Football Throw

  • Thread starter gmunoz18
  • Start date
  • #1
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On this problem i have gotten stuck and i am not really sure why.

Homework Statement



A quarterback claims that he can throw the football a horizontal distance of 169.1 m (185 yd). Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 30° above the horizontal. To evaluate his claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored.

Homework Equations



kinematics equations

The Attempt at a Solution



First I cut the trajectory in half to find the footballs maximum height

than on the y coordinate is at its vertex its velocity at that point is 0 so i was able to put that in as well

So my list came out as is

X coordinate Y coordinate
X=84.55 Y=48.814
Xi=0 Yi=0
a=0 a=-9.8 m/s/s
Vi=? Vi=?
V=? V=0

than in order to get more information i solved for the ball to hit the ground from its highest altitude i.e. 48.814 meters.

dist = 1/2 g t^2 (g=9.8 m/s/s) so i used this equation and came up with 3.1562 seconds

Now I was able to solve for Vi for Y coordinate which i got 30.93m/s

after this i just used the sin(30)=30.93/x and i got 61.8628m/s


Was there an easier way to do this problem am i missing something very simple? but this answer is not matching up and cant seem to get it

thanks i advance
 

Answers and Replies

  • #2
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sorry about the double post!
 
  • #3
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This way may be a little easier, and I got an answer that works.

Just use the kinematic equation:

[tex]\Delta s = v_i t + \frac{1}{2} a t^2[/tex]

for each direction x and y. For each velocity, separate [tex]v_i[/tex] into components. Then you have two equations with two unknowns, so you can then solve for each unknown.
 
  • #4
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would i use the seconds that i solved for t?
 
  • #5
79
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would i use the seconds that i solved for t?

No. You're basically starting over with t as an unknown.

I did the problem myself this way. It's not too bad and really shouldn't take too long, I don't think as long as finding the height then the time the way you did.

To get you started a bit, write down the variables you already know, x, y, a.

Your unknowns are v and t. But you do know how to split v into [tex]v_1x[/tex] and [tex]v_1y[/tex]. So, then you should be able to solve for both t and v.
 
  • #6
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ohh wow that is alot easier thanks for the help
 
  • #7
Its a good practice to go the basic way--splitting the motion into horizontal and vertical, but this question can also be solved by using this formula for horizontal range. Isn't it faster to put in the values and get the result ?

[tex] R=\frac{v^2 sin2\theta}{g}[/tex]
 
  • #8
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Its a good practice to go the basic way--splitting the motion into horizontal and vertical, but this question can also be solved by using this formula for horizontal range. Isn't it faster to put in the values and get the result ?

[tex] R=\frac{v^2 sin2\theta}{g}[/tex]

I have not encountered this equation yet but when i used it i did not get the correct answer. this is as R as the entire range correct?
 
  • #9
I am pretty much confident that this equation is correct. and yes, R is the complete horizontal range, which, according to your question is 169.1 m

I got v=43.74 m/s.

maybe I'm wrong..:redface:
 
  • #10
79
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I am pretty much confident that this equation is correct. and yes, R is the complete horizontal range, which, according to your question is 169.1 m

I got v=43.74 m/s.

maybe I'm wrong..:redface:

No, that's the answer I got doing it the other way, but I'm not familiar with that equation either.
 
  • #11
No, that's the answer I got doing it the other way, but I'm not familiar with that equation either.

Maybe you'll encounter these equations later. They are all over in books on projectile motion.
My teacher derived these equation for us, by splitting the motion into horizontal and vertical components.
Its like this :

Time of flight=[tex] \frac{2vsin\theta}{g}[/tex]

Range= [tex] \frac{v^2 sin2\theta}{g}[/tex]

Max height=[tex]\frac{v^2 sin\theta}{2g}[/tex]
 
  • #12
29
0
I am pretty much confident that this equation is correct. and yes, R is the complete horizontal range, which, according to your question is 169.1 m

I got v=43.74 m/s.

maybe I'm wrong..:redface:

thats what i got i did it with only half the distance at first ha but when i did it for the full 169.1 it was perfect!

Thanks alot googlespider and chococat you helped alot!

What physics are you in google_spider? general, calc based?
 

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