What Is the Minimum Speed Needed for a Boulder to Clear the Dam?

  • Thread starter Thread starter Bottomsouth
  • Start date Start date
  • Tags Tags
    Rolling Speed
Click For Summary
SUMMARY

The minimum speed required for a 76 kg boulder to clear a dam while rolling horizontally off a 20m cliff is calculated to be 45.45 m/s. The calculations involve determining the time of flight using the equation for free fall, leading to the horizontal distance covered. The correct arithmetic confirms that the boulder must maintain this speed to avoid striking the dam located 100m from the cliff's base. The final verification of the calculations indicates that the initial speed is indeed accurate despite initial confusion.

PREREQUISITES
  • Understanding of projectile motion and free fall physics
  • Familiarity with kinematic equations
  • Basic algebra for solving equations
  • Knowledge of gravitational acceleration (9.8 m/s²)
NEXT STEPS
  • Study projectile motion in physics, focusing on horizontal and vertical components
  • Learn to apply kinematic equations in real-world scenarios
  • Explore the effects of air resistance on projectile motion
  • Investigate advanced topics in dynamics, such as energy conservation in projectile motion
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the principles of projectile motion and its applications in real-world scenarios.

Bottomsouth
Messages
27
Reaction score
0

Homework Statement



a 76 kg boulder is rolling horizontally at a top of a vertical cliff that is 20m above the surface of a lake. 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 m below the top of the dam. 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?

Homework Equations





The Attempt at a Solution



20= 4.9 * t^2
20/4.9= t^2
t=square 20/49
t=2.02

100 = vx0 * t
100/2.02 = vx0
vx0 = 45.45 m/s

20+25 = .5 * 9.8 * t^2
45 = 4.9* t^2
t+3.03

V0x * 3.3 = 150m
45.45*3.3 = 150
150-100 = 50m

I am not getting the min speed. Is it the 45.45? I believe so, but the program doesnt. Appreciate any help. The 50 m is correct.
 
Physics news on Phys.org
Bottomsouth said:
20= 4.9 * t^2
20/4.9= t^2
t=square 20/49
t=2.02

100 = vx0 * t
100/2.02 = vx0
vx0 = 45.45 m/s
Redo that last bit of arithmetic.
 
49.5, thanks. Have had very little sleep.
 

Similar threads

Boulder rolling off of cliff: will it hit the village?
  • · Replies 16 ·
Replies
16
Views
3K
Projectile motion with mass height and range
  • · Replies 2 ·
Replies
2
Views
2K
Projectile Motion Off a Cliff: Solving for Minimum Speed and Distance Traveled
  • · Replies 8 ·
Replies
8
Views
11K
Minimum Speed for a Horizontally Projected Diver to Clear a Ledge
  • · Replies 7 ·
Replies
7
Views
2K
Projectile Motion: Boulder from Cliff to Plain
  • · Replies 1 ·
Replies
1
Views
4K
What is the Minimum Speed for a Diver to Clear a Cliff Ledge?
  • · Replies 3 ·
Replies
3
Views
8K
What is the minimum speed required for Spiderman to catch a falling baby?
  • · Replies 4 ·
Replies
4
Views
6K
Sphere Rolling Down Hill into Projectile Motion
  • · Replies 3 ·
Replies
3
Views
2K
What's the minimum initial speed to score
  • · Replies 6 ·
Replies
6
Views
2K
NEED PHYSICS HELP am i doing this correctly?
  • · Replies 6 ·
Replies
6
Views
3K