How high is the snow pile if a kid jumps off it and lands on the ground?

  • Thread starter Thread starter Drizzy
  • Start date Start date
AI Thread Summary
The discussion centers on calculating the height of a snow pile based on a kid jumping off it and landing on the ground. The key equation used is S = v0t + (at^2)/2, with the assumption that the kid's highest point occurs when velocity is zero at 0.4 seconds. The calculations suggest that the snow pile's height is derived from the total time of 0.8 seconds for the jump. However, a misunderstanding arises regarding the sign of the gravitational acceleration and the interpretation of the velocity at the peak of the jump. Clarification is needed on the relationship between the velocity graph and the snow pile's height in relation to the ground level.
Drizzy
Messages
210
Reaction score
1

Homework Statement


The diagram shows a kid jumping off a snow pile into the grown that is also covered with snow. So the kid is jumping up and then lands on the ground.
https://scontent-arn2-1.xx.fbcdn.net/hphotos-xpf1/v/t34.0-12/12380374_1234806623215387_1066477107_n.jpg?oh=fe84392440a1ddbf0eebf557109d56ec&oe=5677CAA8
The question is: how tall is the snow pile if the ground is 0 m?

Homework Equations



S=v0t + (at^2)/2

The Attempt at a Solution


I am thinking that the kid is at its highest point when the velocity is zero and t=0.4. It takes 0.4 seconds up and also 0.4 seconds down to the level where the snow pile was at. so the snow piles height is from 0.8 seconds to 1,2 seconds. 1.2-0.8=0.4
s= -4*0.4 + (9.82 * 0,4^2)/2 = - 0,8144

What went wrong in my assumptions or solution?

upload_2015-12-19_14-42-36.png
 
Physics news on Phys.org
If it's headed downwards, you are denoting this as a negative velocity, so g in your equation will likewise be negative.
 
Also, when the velocity is at its most negative point, that's when the kid reaches the top of the layer of snow that's on the ground, not the ground itself. Ponder the finish of the velocity vs time graph.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top