Solving Basic Vector Problems: Understanding Direction and Speed in Cricket

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A cricketer running north at 10 m/s throws a ball west at 24 m/s, leading to a combined velocity that can be analyzed using vector addition. The resulting speed of the ball is calculated to be 26 m/s using the Pythagorean theorem, as it forms a right triangle with the north and west components. The direction of the ball is determined to be 67 degrees 23 minutes west of north, calculated using the arctangent of the ratio of the westward to northward speeds. The discussion emphasizes the importance of correctly visualizing the triangle formed by the vectors to arrive at the correct angle. Understanding these vector components is crucial for solving similar problems in physics.
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This is problem very basic for all of you but I am struggling to grasp this.

Heres the problem
A cricketer is running due north with a cricket ball at 10 m/s and throws the ball due west at 24 m/s. Find the direction of the ball and its speed.

Please give me a details explanation as I am completely 100% lost with this.
 
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You can define two vectors given your data, say ##\vec u## and ##\vec v##:
  1. Can you find ##\vec u+\vec v## ?
  2. What does the vector ##\vec u+\vec v## represent?
 
Worked out the speed 26 m/s Answer sheet tells me 67 degrees 23 min west of north? How does it arrive at that??
 
tinybang said:
Answer sheet tells me 67 degrees 23 min west of north? How does it arrive at that??

Draw the triangle that is made up of the (final) velocity vector and it's Northern and Western components.

Can you use the known information to find the direction?
 
Did you understand what Hakim Philo said? The cricketer is running N at 10 m/s. You can take his velocity vector as a vector with length 10 pointing N. The ball, which was moving along with the cricketer, is thrown W at 24 m/s. You can take its velocity vector, relative to the cricketer, to be a vector with length 24 pointing W. Since N and W are at right angles, their vector sum is the hypotenuse of a right triangle with legs of length 10 and 24. You say you got 26 m/s for the speed of the ball, relative to the ground. I presume you got that using the Pythagorean theorem. The angle can be found by using the fact that if a right triangle has "opposite side" of length a and "near side" of length b, then the angle is given by tan(\theta)= a/b so that \theta= arctan(a/b).
 
ahh got it. Thanks for the help. Turns out I drew the triangle upside down so of course I was never going to get the degrees correct. :)
 
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