Solving Projectile Problem: Initial Velocity from Cliff

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SUMMARY

The projectile problem involves calculating the initial velocity (Vi) of a cannon fired at a 30-degree angle from a 20m high cliff, landing 40m away from the base. The relevant equations include the horizontal motion equation Vi cos(30°)t = 40 and the gravitational acceleration g = 9.8 m/s². To solve for Vi, one must determine the total time of flight (t) and apply it to both horizontal and vertical motion equations. The discussion emphasizes the need for a second equation to account for vertical displacement during the projectile's flight.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions, specifically cosine
  • Knowledge of kinematic equations for horizontal and vertical motion
  • Basic grasp of gravitational acceleration (g = 9.8 m/s²)
NEXT STEPS
  • Study the derivation of projectile motion equations
  • Learn how to apply kinematic equations in two dimensions
  • Explore the concept of time of flight in projectile motion
  • Investigate the effects of angle and height on projectile trajectories
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Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators seeking to clarify concepts related to initial velocity calculations in projectile problems.

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Homework Statement



A cannon is fired at an angle of 30 degrees above the horizontal from a cliff that is 20m above a flat river bottom. What is the initial speed of the projectile if it is found to land 40m from the base of the cliff?



Homework Equations


g = 9.8ms^-2
Vi = initial velocity

The Attempt at a Solution


I know that Vi cos 30t = 40, but does the "t" represent the time taken for the entire projectile motion, or just the upper symmetrical parabolic trajectory? I'm stuck at Vi cos 30t = 40 because what I did after simply didn't make any sense at all.

I've been stuck on this question for days on end, please help me with it! Thanks! :D
 
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The t represents the entire time of flight, from your equation you can actually solve for it. Now how about a second equation for the vertical part of the flight?
 

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