Solving Range & Height of Spring Gun Ball - 45° Angle

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SUMMARY

The discussion centers on solving projectile motion problems involving a spring gun that launches a ball at a 45-degree angle. The ball achieves a horizontal range of 32 feet, and the maximum height needs to be calculated. Additionally, the discussion addresses finding the two angles that yield a range of 20 feet with the same initial velocity. Key parameters include gravitational acceleration at -9.8 m/s² and initial conditions such as Xo = 0 and Yo = 0.

PREREQUISITES
  • Understanding of projectile motion principles
  • Knowledge of kinematic equations
  • Familiarity with trigonometric functions
  • Basic grasp of gravitational acceleration (9.8 m/s²)
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  • Calculate maximum height using the formula for projectile motion
  • Explore the derivation of projectile motion equations
  • Learn about the effects of launch angle on range
  • Investigate the relationship between initial velocity and range
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Students in physics, educators teaching projectile motion, and anyone interested in understanding the dynamics of spring guns and projectile trajectories.

physics dud
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I am not sure I can solve this, I am confused that I do not have the information I need. I am lost and my prof is of little help.

A spring gun projects a ball at an angle of 45 degrees above the horizontal. The ball has a horizontal range of 32 feet. What is the maximum height to which the ball rises, and for the same initial velocity, what are the two angles for which the range is 20 feet.

Dont I need at least a time frame to solve this?

The best part is, the other two problems are even more vague to me.
 
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Draw a picture of what is going on in the problem, then write down what you know and show in words what you've done. That way if you made a mistake someone can help you. It may also be helpful to write down what you know and write down what you don't know, then you know what to solve for
 
Xo = 0
Yo = 0
Ax = 0
Ay = -9.8m/s
Angle Theta = 45
x = 32 feet ( 9.6 m, 960 centimeters)
 

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