How Does Cannonball Trajectory Change with Height?

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SUMMARY

The discussion focuses on the physics of cannonball trajectories, specifically analyzing a scenario where a cannon is fired vertically from the base of an 85 m cliff and another cannon is fired horizontally from the top. The initial speed required for the vertical cannonball to reach the cliff's height is calculated to be 40.8375 m/s. The participants explore the relationship between vertical and horizontal projectile motion, emphasizing the need for equations that relate to range, particularly for the horizontal cannon fired from the cliff's summit.

PREREQUISITES
  • Understanding of kinematic equations, specifically v^2 = 2gh
  • Knowledge of projectile motion concepts, including horizontal and vertical components
  • Familiarity with the maximum range formula for projectile motion, ymax = (v^2 sin^2 ∅)/2g
  • Basic algebra skills for solving equations
NEXT STEPS
  • Research the derivation of the range formula for projectile motion
  • Study the effects of initial velocity and angle on projectile range
  • Explore numerical methods for validating projectile motion calculations
  • Learn about the impact of air resistance on projectile trajectories
USEFUL FOR

Students studying physics, educators teaching projectile motion, and anyone interested in understanding the dynamics of cannonball trajectories in relation to height and range.

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Homework Statement


A cannon is placed at the bottom of a cliff 85 m high. If the cannon is fired straight upward, the cannonball just the reaches the top of the cliff.
a) Calculate the initial speed of the cannonball.
b) Suppose a second cannon is placed at the top of the cliff and fired horizontally with the same initial speed as part (a). Prove numerically that the range of this cannon is the same as the maximum range of the cannon from the base of the cliff.


Homework Equations


for a) v^2 = 2gh
for b) ymax = (v^2 sin^2 ∅)/2g


The Attempt at a Solution



for a) i got v = 40.8375 m/s

for b) i got ymax = 85 meters but that is only for the vertical part.

I am not sure how to find the numerically part
 
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popmop2 said:
for b) ymax = (v^2 sin^2 ∅)/2g
Not a useful equation for this question. Do you have any equations relating to range?
 

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