Where Does a Projectile Fired from a Cliff Land?

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SUMMARY

A projectile is launched from a 200-meter cliff with an initial velocity of 60 m/s at an angle of 60 degrees above the horizontal. The time of flight calculated is 6.39 seconds, leading to a horizontal distance of 191.7 meters. However, the expected result is 0.41 kilometers, indicating a discrepancy in the calculations. The vertical component of the initial velocity, calculated as 60sin(60), is crucial for determining the correct trajectory and landing distance.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions, specifically sine
  • Knowledge of kinematic equations for horizontal and vertical motion
  • Basic grasp of gravitational acceleration (g = 9.81 m/s²)
NEXT STEPS
  • Review the derivation of the projectile motion equations
  • Learn how to calculate the vertical and horizontal components of velocity
  • Study the effects of initial height on projectile landing distance
  • Practice solving similar projectile motion problems with varying angles and heights
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators looking for examples of problem-solving in kinematics.

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Homework Statement


A projectile is fired into the air from the top of a 200-m cliff above a valley. It's initial velocity is 60 m/s at 60 degrees above the horizontal. Where does the projectile land?


Homework Equations


delta x = intial velocity of x * t
delta y = -.5*g*t

The Attempt at a Solution


I solved for time and got 6.39s. When I put this back in for delta x I get 191.7m. However according to my professor the results should be .41km. Any ideas? Thanks!
 
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You have an initial speed in the y direction, 60sin60.
 
Thanks.
 

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