Projectile Motion: Boulder from Cliff to Plain

Click For Summary
SUMMARY

The discussion focuses on calculating the minimum speed required for a 76.0 kg boulder to clear a 100 m horizontal distance from a cliff that is 20.0 m high, landing on a plain 25.0 m below the dam. The key equations involved are the equations of motion and projectile motion principles. It is established that the mass of the boulder does not affect the horizontal motion, allowing for the calculation of the required speed without needing to incorporate mass into the equations.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with equations of motion
  • Basic knowledge of kinematics
  • Ability to analyze horizontal and vertical motion separately
NEXT STEPS
  • Study the equations of motion in detail
  • Learn how to apply projectile motion equations to real-world problems
  • Explore the concept of horizontal and vertical components of motion
  • Investigate the effects of gravity on projectile trajectories
USEFUL FOR

Students studying physics, educators teaching projectile motion concepts, and anyone interested in solving real-world motion problems involving gravity and horizontal distance.

jackleyt
Messages
20
Reaction score
0
1. A 76.0 kg boulder is rolling horizontally at the top of a vertical cliff that is 20.0 m above the surface of a lake, as shown in figure below. (Intro 1 figure) The top of the vertical face of a dam is located 100 m from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is 25.0 m below the top of the dam.

1)What must the minimum speed of the rock be just as it leaves the cliff so that it will travel to the plain without striking the dam?

2)How far from the foot of the dam does the rock hit the plain?


Homework Equations


Equations of Motion, Projectile Motion


The Attempt at a Solution


I don't know how to incorporate the mass of the rock in the equation. I don't know where to start!
 
Physics news on Phys.org
I don't know how to incorporate the mass of the rock in the equation. I don't know where to start!
Perhaps you don't need the mass? How would you approach the problem if you didn't know what the mass was?

Sometimes you don't always use all the information given to you in the question.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
12K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
13
Views
4K
Replies
2
Views
10K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K