Golf ball problem (velocity, max height, hang time)

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SUMMARY

The discussion centers on solving the "golf ball problem," which involves calculating the velocity, maximum height, and hang time of a golf ball in projectile motion. The initial velocities are established as Vx = 20 m/s and Vy = 34.6 m/s. Key formulas include the displacement equations for both x and y directions: x(t) = x₀ + v₀x*t and y(t) = y₀ + v₀y*t - 0.5*g*t², where g = 9.8 m/s². The user is guided to apply these formulas to find maximum height and hang time effectively.

PREREQUISITES
  • Understanding of basic physics concepts related to projectile motion
  • Familiarity with kinematic equations
  • Knowledge of vector components in motion
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Learn how to calculate maximum height using the formula h = (Vy²)/(2g)
  • Explore the concept of hang time and how to derive it from initial velocities
  • Investigate the effects of air resistance on projectile motion
USEFUL FOR

Students studying physics, educators teaching projectile motion, and anyone interested in understanding the dynamics of objects in motion.

aquaorange
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"golf ball problem" (velocity, max height, hang time)

Okay. I'm having a really hard time with this. I know how to start the problem, but after that I go nuts, and literally start pulling out my hair. There is no book I can look in for reference, as this part is teacher supplemented.

The given picture/information is this:

physics-question.jpg


And I need to find:

velocity of x at zero time:
velocity of y at zero time:
maximum height (from ground, not cliff):
hang time:
velocity of x after "a" time:
velocity of y after "a" time:

Now, I can find Vx at zero (20m/s), and Vy at zero (34.6m/s). I also know that Vx at any time is going to remain the same (20m/s), but then I'm lost for maximum height, hang time, and Vy at "a" time. Gravity is taken into account, but not air resistance. I know that it loses -9.8m/s going up due to gravity, and +9.8m/s going down due to gravity. And I tried doing that a long way, subtracting 9.8 from 34.6, subtracting 9.8 form that answer, etc., and counting how many times to get a seconds number, but then I didn't know how to figure out when I got down to 5.2 and subtracting 9.8 from that. I really have no clue how far up it goes, or how to find anything past making the triangle to find the forces using sine/cosine/tangent.

I'm not sure if there is a formula or multiple formulas to figure this information out. That's what I think I'm missing to get the rest of the information.
 
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The formula which will be of great use to you is the displacement formula. For the x direction, you have x(t) = x_{0} + v_{0x}t, and, for the y direction, y(t) = y_{0} + v_{0y}t - \frac{1}{2}gt^2. You should be able to solve the problem now.
 

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