Solving for Angle and Distance in Newton Mechanics Course | 2 Men on Icy Lake

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In summary: I can't follow how you proceeded in (c). Did you use K.E.? But using conservation of momentum I ended up with the same algebraic expression as you have for their speed after catching the ball.In summary, the two men threw the balls at the same time and at an angle of 14.17° to each other, with a speed of 10.0 m/s after they caught the balls.
  • #1
ToHHoR
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Homework Statement


Relevante for the problem::
- We are taking a course in Newton mechanicsTwo men, each of mass M = 75.0 kg find themselves standing on an icy lake, a distance d = 5.00 m apart. They each happen to carry a basketball, weighing m = 0.600 kg. In order to move, they get the clever idea of throwing the balls to each other. They throw with speed v = 10.0 m/s (both of them). Assume they throw at the same time.

a) With what angle should they throw, in order to hit the other guy? b) How far are the two guys apart, when they receive the balls?
c) What is their speeds after they receive the balls?

For each question, provide an algebraic expression, as well as a numerical result with appropriate units.
upload_2016-11-4_19-44-9.png

Homework Equations


Range equ. : D = (v(ball) * sin 2 theta)/ m
D = v(person) * t + d

The Attempt at a Solution


My idea was to calculate the time when the ball is in the air. For then to see how far the persons have traveled back before receiving the ball again. For then to use the range equ. to find the angle.
 
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  • #2
Welcome to PF!

As you may be aware, we can only provide hints to help you solve your problem. In order to do that we need to see your work.

Can you show us what you did to solve your problem and where you got stuck or got the wrong answer?
 
  • #3
screenshot1.jpg
Hi, trying hard to get this picture out but hoping this will work.

All we want to know is; can we solve it the way we are thinking?

https://goo.gl/photos/pQZLW1HsU8inWVGZ6

[ Mod Note: google photos offers no direct link, so had to take a screenshot and have attached that ]
 
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  • #4
:welcome:

Your eqn. for Vp2 is right.

Then you use but don't show the origin of the eqn.
##d\ =\ \frac{10.sin 2\theta}{m_b}\ \ ## Where did this come from?

Does your textbook provide the correct answers, so I can check my result?
 
  • #5
Thank you!

The eqn. is the range eqn, for a flat surface. It was wrong, it should look like d = (vi2sin2*thetha)/mb

We don't have an answer..
 
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  • #6
ToHHoR said:
The eqn. is the range eqn, for a flat surface. It was wrong, it should look like d = (vi2sin2*thetha)/mb
An improvement, but I still don't like it. Have you looked at the units of this expression? Are you quoting it from somewhere or have you derived it and made a mistake because I'm sure that gravity should feature in any expression for range.
 
  • #7
An improvement, but I still don't like it. Have you looked at the units of this expression? Are you quoting it from somewhere or have you derived it and made a mistake because I'm sure that gravity should feature in any expression for range.

Yes, NascentOxygen, it should be g, not mb.Think I worked out the solution now...

If I take 90 - 14,8 = 85,2.. is this the second solution?

IMG_20161110_124943.jpg
IMG_20161110_142728.jpg
 
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  • #8
If I take 90 - 14,8 = 75,2.. is this a second solution?
 
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  • #9
We have agreement on the expression for ##sin (2\theta)## but I evaluated ##\theta## to be 14.17°

I can't follow how you proceeded in (c). Did you use K.E.? But using conservation of momentum I ended up with the same algebraic expression as you have for their speed after catching the ball.
 
  • #10
ToHHoR said:
If I take 90 - 14,8 = 75,2.. is this a second solution?
There often is a second parabolic path to consider in these types of questions. You could test your "guess", though guesses rarely turn out right.

Instead, go back to your expression for ##2\theta## and find another ##2\theta## angle having that same value for sine. Then test it.
 

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