Water Balloon Drop Final Exam Review

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SUMMARY

The discussion centers on calculating the height from which a water balloon must be dropped to hit a target 3.0 meters away, considering a student walking speed of 1.5 m/s. The physics equation used is yf=yi+Vi(delta T)+(1/2)g(delta T)^2, leading to the conclusion that the water balloon must be dropped from a height of 19.6 meters. The time calculated for the balloon to reach the target is 2 seconds, confirming the accuracy of the height determination. This solution effectively combines kinematic equations with real-world application.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Basic knowledge of gravitational acceleration (g = 9.8 m/s²)
  • Ability to calculate time using distance and speed
  • Familiarity with projectile motion concepts
NEXT STEPS
  • Review the derivation of kinematic equations in physics
  • Learn about projectile motion and its applications
  • Explore real-world examples of physics in sports and engineering
  • Practice solving problems involving free fall and horizontal motion
USEFUL FOR

Students studying physics, educators preparing exam materials, and anyone interested in applying physics concepts to practical scenarios.

charan1
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Im doing Final exam review this is the first question please check! Thank you!

Homework Statement


1. Water balloon drop. You and your physics major roommate devise a plan to drop water balloons on
classmates entering your dorm. Just as they cross a mark that’s 3.0 m from the door, you’ll drop a water
balloon. How high above the classmates entering the front door do you need to be to carry out your plan?
You’ve determined that most students walk at a speed of about 1.5 m/s.


Homework Equations


yf=yi+Vi(delta T)+(1/2)g(delta T)^2

(delta x)/ T = Velocity

The Attempt at a Solution



3 meters / 1.5 m/s = 2 seconds = T

yf=0 ; yi=? ; Vi=0 ; Delta T = 2 s

0=yi+(0)(2)+(.5)g(2)^2

yi=19.6m

To drop the balloons the students have to be 19.6 meters high to hit their target!
 
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