Projectile Motion Analysis: Calculating Safe Distance for a Swimmer in Training

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SUMMARY

The discussion centers on calculating the safe distance for a swimmer training below a projectile fired from a plateau. The projectile is launched from a height of 150m with an initial velocity of 320 m/s. The correct angle of elevation is determined to be approximately 41.4 degrees, leading to a calculated range of about 6.8 km. Therefore, the swimmer must maintain a distance of at least 6.8 km from the cliff to ensure safety from the projectile.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with the range formula R=v0^2*sin(2*theta)/g
  • Knowledge of maximum height calculation h=v0^2*sin^2(theta)/2g
  • Basic grasp of trigonometry and angles
NEXT STEPS
  • Study the derivation of the projectile motion equations
  • Learn about the effects of air resistance on projectile motion
  • Explore real-world applications of projectile motion in sports
  • Investigate advanced topics like optimization of launch angles for maximum distance
USEFUL FOR

Physics students, educators, athletes in training, and anyone interested in the practical applications of projectile motion analysis.

synchronicity
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Hey there, fellow physics fans!

i was wondering if anyone is well versed in the language of projectile motion-

i did this problem in which a projectile is fired from a plateau at a height of 150m above sea level, at a velocity of 320 m/s. it is fired from a distance of 8km away from the edge of a cliff that drops down 150m to the water. some crazy swimmer happens to be down there in the water training for the olympics, and the problem is to figure out how far away from the cliff he'll have to be so that he doesn't get bonked in the head by that projectile. using the height above the water and the initial angle of elevation determined from R=v0^2*sin(2*theta)/g, i keep coming up with a distance of approximately 300m, but I'm not sure if this is correct. what do you guys and gals think?

merci!

synchronicity
 
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Hi there! Projectile motion can definitely be a tricky concept, but let's see if we can help you out with this problem.

First, let's break down the information we have been given. We know that the projectile is fired from a height of 150m above sea level, with an initial velocity of 320 m/s. The cliff is 150m high and the swimmer is training in the water below. We also have the formula for range, R=v0^2*sin(2*theta)/g, where v0 is the initial velocity, theta is the angle of elevation, and g is the acceleration due to gravity.

To solve this problem, we need to find the angle of elevation at which the projectile is fired. We can do this by using the formula for maximum height, h=v0^2*sin^2(theta)/2g, and setting it equal to the height of the plateau (150m). This will give us the angle of elevation as approximately 41.4 degrees.

Now, using this angle and the range formula, we can calculate the horizontal distance the projectile will travel. Plugging in the given values, we get a range of approximately 6.8 km. This means that the swimmer should be at least 6.8 km away from the cliff in order to avoid getting hit by the projectile.

I hope this helps! Remember, always double check your calculations and keep in mind the units of measurement. Good luck with your training!
 

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