How High Was the Shelf If a Ball Rolls Off at 8m/s and Hits the Ground in 0.22s?

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion, specifically analyzing the height of a shelf from which a ball rolls off horizontally at a speed of 8 m/s and takes 0.22 seconds to hit the ground.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to determine the height of the shelf, with some focusing on vertical motion while others consider horizontal distance. There are questions about the necessity of calculating both vertical and horizontal distances, and participants explore the implications of initial velocities and acceleration due to gravity.

Discussion Status

The conversation includes various attempts to apply kinematic equations, with some participants correcting each other's calculations and questioning the setup of the problem. There is an ongoing exploration of the relationship between vertical and horizontal components of motion, but no explicit consensus has been reached on the final answer.

Contextual Notes

Participants note the importance of using the correct initial conditions and values in their calculations, with some confusion arising from the different distances calculated and the assumptions made about the motion of the ball.

PhilMTSU12
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Homework Statement


A ball rolls off a table with a horizontal v of 8m/s. If it takes the ball .22s to hit the floor, how high was the shelf?


Homework Equations


d=volt+1/2at²


The Attempt at a Solution


HORIZ
a=0
Vo=8m/s
V=8m/s
d=?
t=.22s

d= (8)(.22)+1/2(8)(.22²)

Is the hieght of the shelf d=1.997m?
 
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You have to do it all in the Y-direction.

Vo = 0
t = 0.22
a = gravity

d=volt+1/2at²

solve for d
 
correction d= (8)(.22)+1/2(0)(.22²)
d= (8)(.22)
d= 1.76
 
ok Vertical
a= -9.8m/s
Vo=0
V=?
d=9.8m
t=.22

(9.8)(.22)+1/2(9.8)(.22²)
d= 2.39
 
PhilMTSU12 said:
ok Vertical
a= -9.8m/s
Vo=0
V=?
d=9.8m
t=.22

(9.8)(.22)+1/2(9.8)(.22²)
d= 2.39

Recheck your working, you made a mistake

Your equation there is:

at+1/2at2 instead of volt+1/2at2

its just 1/2(9.8)(.222)
 
ok vert= .24 and horz=1.76

what do I do with these 2 distances
 
The vertical distance is the height of the shelf
 
so, I only needed to find the vertical distance? I did the horz distance for no reason...

and why do i only need half of the equation?
 
You didn't really only need half, but since Vo = 0, then volt = 0

so volt+1/2at2 = 0 + 1/2at2
 
  • #10
ok...
The next problem is this

A ball rolls off a table which is 2.8m high. If it leaves the table with a horizontal v of 8m/s, calc the distance from the base of the shelf at which the ball will strike the ground.

I know i need to figure out how long it will take the ball to strike the ground. then use that to find the horz distance

vert d= 1.41

8(1.41)= 11.28m
 
  • #11
You're theory is right, but your answer confuses me, what is that?

How are you finding the time the ball takes to hit the ground?
 
  • #12
i tried to quickly do all the work haha

vert distance
a= -9.8m/s
Vo=0
V=?
d=9.8m
t=?

using at+1/2at²

9.8= 1/2(9.8)t²

2(9.8)=9.8t²
over
9.8=9.8
(the 9.8's cancel leaving)

2=t²

t=square root of 2

t= 1.41


soo it will take 1.41s for the ball to hit the ground.

using t= 1.41 i can now find the horz distace

a=0
Vo=8
V=8
d=?
t=1.41

d= volt

8(1.41)= 11.28m
 
  • #13
Well, your method of working is correct, but in the very beginning you used the wrong value of d

"A ball rolls off a table which is 2.8m high" not 9.8 like you used
 
  • #14
ok

2.8m= 1/2(9.8)t²

2(2.8)=2(1/2)(9.8)t²

5.6m= 9.8t²
Divide 9.8 to both sides

.5714=t²

t=.756s

d=volt
d=8(.756)
d=6.048
 
  • #15
Does the final anwser look correct?
 

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