What is the Path of a Cricket Ball During a Bowler's Run Up?

  • Thread starter Thread starter eejpama
  • Start date Start date
  • Tags Tags
    Ball Path
AI Thread Summary
The discussion focuses on modeling the path of a cricket ball during a bowler's run-up and the physics involved in its trajectory. Participants emphasize the importance of understanding that a bowler does not "throw" the ball but bowls it with a straight arm, which is crucial for accurately modeling its motion. The conversation explores the effects of bouncing on the ground, noting that the ball's velocity just after the bounce is not zero but changes direction. Suggestions include modeling the impact as either an elastic or inelastic collision to better understand the ball's behavior upon bouncing. Overall, the thread provides insights into the physics of cricket ball dynamics during a bowler's delivery.
eejpama
Messages
3
Reaction score
0

Homework Statement


model the run up a bowler takes before he throws the cricket ball


Homework Equations


model the path of the cricket ball


The Attempt at a Solution


we can make any assumption. if anyone good in physics please helo me model it.
 
Physics news on Phys.org
If this is homework you should have a go yourself first.

PS: He doesn't just "throw" the cricket ball (that would be an illegal delivery) he bowls it with a straight arm. I'm not just being picky. This might help you answer the question.
 
i have done some work and equations. but when balls hits the ground i don't understand how to work it out because bounce on ground before it rise again to touch the bat. at that moment i want to know the initial velcity whether it is zero or not? just confused
 
No not zero velocity. If it bounces the [STRIKE]velocity[/STRIKE] speed just after the bounce won't be very different to that just before the bounce. Just the direction changes.

You could model the impact with the ground as an elastic collision or an inelastic collision or perhaps just assume the speed doesn't change. Although obviously the direction does (up rather than down).

http://en.wikipedia.org/wiki/Elastic_collision

"An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter" (eg none lost to heat, sound etc)

http://en.wikipedia.org/wiki/Inelastic_collision
"Inelastic collisions may not conserve kinetic energy, but they do obey conservation of momentum"
 
thank u that is really helpfull.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top