Solving Steve and Mark's Backyard Cricket Problem

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SUMMARY

The discussion focuses on solving a physics problem involving projectile motion in a backyard cricket scenario. Steve bowls a ball at a velocity of 60 km/h from a height of 2.1 meters at an angle of 5 degrees below the horizontal. The key questions are determining where the ball bounces and its speed just before impact. The initial calculations yielded a bounce distance of 4.8 meters and a speed of 19.3 m/s, which the user questions due to the complexity of the problem involving different launch angles.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of vector components in physics
  • Basic skills in algebra and trigonometry
NEXT STEPS
  • Study the kinematic equations for projectile motion
  • Learn how to resolve vectors into horizontal and vertical components
  • Explore the effects of initial height on projectile trajectories
  • Practice solving problems involving different launch angles
USEFUL FOR

This discussion is beneficial for first-year physics students, educators teaching projectile motion, and anyone interested in applying physics concepts to real-world scenarios like sports.

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Hey this is my second post on this subject sorry to bother but i still just can't get my head around it...i have used so many different equations and the answers just don't seem to be correct.

The question is

Steve and Mark are playing a game of backyard cricket. They set their field up so that the batsman stands 10.5metres from the point where the bowler releases the ball. Steve bowls the first ball of the match and the ball leaves his hand with a velocity of 60 km/h at an angle of 5 degrees below the horizontal. The ball is relased from a height of 2.1metres above the ground.

a) Where does the ball bounce
b) What is the speed of the ball just before it bounces?

I got 4.8 metres distance which just doesn't seem likely to me and 19.3 m/s

The next part of the question just changes the angle to 3 degrees above horizontal and asks the same questions, this bit is really throwing me off because all of my textbooks only have equations 'from origin' and the ball leaves 2.1 meters above the ground and goes up then down to 0 meters.

If someone could please explain this to me very simply it would be very greatly appreciated...this is my first year of physics :)
 
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