Solving for Initial Velocity in Projectile Motion

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SUMMARY

The discussion focuses on calculating the initial velocity required for a projectile to achieve a maximum horizontal distance of 136 meters on a level field, with gravity set at 9.8 m/s². The user initially attempted to use the range equation but encountered difficulties in determining the correct angle for maximum range. The optimal angle for maximum range is established as 45 degrees, leading to the conclusion that the initial velocity (vi) is approximately 30.70 m/s when applied to vertical motion.

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  • Understanding of projectile motion principles
  • Familiarity with the range equation for projectile motion
  • Knowledge of trigonometric functions, specifically sine
  • Basic algebra for solving equations
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Students studying physics, educators teaching projectile motion concepts, and anyone interested in the mathematical principles behind throwing objects in sports or engineering applications.

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A man can throw a ball a maximum horizontal distance of 136 m on a level field. The acceleration of gravity is 9.8 m/s^2. How far can he throw the same ball vertically upward from the ground? (Assume that the ball is thrown from the ground and that his muscles give the ball the same speed in each case.) Answer in units of m.

My work: I was trying to use the range equation to solve for vi so 136=v^2sin(2theta)/g ok so 136x9.8= 1332.8=v^2sin(2theta) , I am obviously missing something what is it? thanks
 
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What is the angle [itex]\theta[/itex] for maximum range? Assume no air resistance.

When you obtain the velocity, then apply it to the vertical case.
 
45 degrees and sin(45) is sqrt(2)/2 so really its 1332.8=v^2(sqrt(2)) , which for v i got 30.69905402 thank you, i didnt think of that
 

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