Maximizing Golf Ball Distance and Hole in One: Intro Physics HW Help

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SUMMARY

The discussion focuses on calculating the time a golf ball spends in the air and the maximum distance it can travel when hit at a speed of 27.0 m/s. The optimal launch angle for maximizing distance in projectile motion is confirmed to be 45 degrees. The equations of motion used include the vertical and horizontal displacement equations, which are simplified to derive the time of flight and range. The final solution for the longest hole-in-one distance is determined using the range equation, R = V₀x * T.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of differentiation in calculus
  • Ability to express initial velocity components (V₀y and V₀x)
NEXT STEPS
  • Study the derivation of the range formula for projectile motion
  • Learn about the effects of launch angle on projectile distance
  • Explore the concept of air resistance in projectile motion
  • Investigate advanced kinematic equations for varying launch heights
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Students studying physics, golf enthusiasts interested in optimizing their game, and educators teaching projectile motion concepts.

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hey everyone, i have a few questions that i just don't understand.
1) a golfer imparts a speed of 27.0m/s to a ball, it travels the max possible distance b4 landing on the green. the tee and the green are at the same level. a) how much time does the ball spend in the air? b) wat is the longest "hole in one" that the golfer can make, if the ball does not roll when it hits the ground?


i would really really appreciate ur help and if someone can do these problems can u please give clear solutions?
thanx a lot
 
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I believe 45 degrees is the angle that produces the longest distance in projectile motion.
 
a)
y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2
x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2
Both equations (especially the latter) reduce quite nicely with some logical thinking (what sort of velocity has an effect on the time?). Then just put the equations together and find the maximum (hint: derivate). Quickly done: 45 degrees does look like the correct answer. Now just throw the angle into the (reduced) equation you got in the first step.
Hint: Express v_0 in v_{0y} and v_{0x}

b) You already got the equation you need in a). Now just throw the velocity in and you're done.
 
Last edited:
Päällikkö said:
a)
y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2
x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2
Both equations (especially the latter) reduce quite nicely with some logical thinking (what sort of velocity has an effect on the time?). Then just put the equations together and find the maximum (hint: derivate). Quickly done: 45 degrees does look like the correct answer. Now just throw the angle into the (reduced) equation you got in the first step.
Hint: Express v_0 in v_{0y} and v_{0x}

b) You already got the equation you need in a). Now just throw the velocity in and you're done.

thanks that also just helped me
 
but i simply used the equation for range to find longest
distance which is same as X, R=X= Vox*T,
 

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