Solve the Tough Question: Roof Tile Falling from a Building

  • Thread starter Thread starter nofear001
  • Start date Start date
Click For Summary
SUMMARY

A roof tile falls from rest from the top of a building, taking 0.20 seconds to pass a 1.6-meter window. The acceleration due to gravity is 9.81 m/s². The initial velocity of the tile as it passes the top of the window is calculated to be 6.038 m/s. Using the equations of motion, the distance from the roof to the top of the window is determined to be approximately 1.86 meters.

PREREQUISITES
  • Understanding of uniform acceleration equations
  • Knowledge of kinematic equations in physics
  • Familiarity with basic calculus for substitution methods
  • Ability to perform calculations involving gravitational acceleration
NEXT STEPS
  • Study the kinematic equations of motion in detail
  • Learn how to derive equations involving initial and final velocities
  • Explore advanced problems involving free fall and projectile motion
  • Practice solving physics problems using substitution methods
USEFUL FOR

Physics students, educators, and anyone interested in understanding the principles of motion and gravity in real-world scenarios.

nofear001
Messages
3
Reaction score
0
Ive been working on this and its a little tough: Whats the answer?

A roof tile falls from rest from the top of a building. An observer inside the building notices that it takes 0.20s for the tile to pass her window whose size is 1.6m. How far above the top of this window is the roof?
(Hint: 2 calculations- Use uniform acceleration equations)

How am i suppose to do this?

a=9.81m/s^2 (im in Canada we use 9.81 not 9.8)
t= 0.20s
d= ?
Vi= 0m/s
Vf=?

Thxs
 
Physics news on Phys.org
Hint: the initial velocity (that is, the velocity as it passes the top edge of the window) is not zero. You need to solve for this velocity, as it will tell you how far from the roof the top of the window is.

- Warren
 
but the rooftile falls from rest rest= 0m/s
 
is the answer 2.51 meters?
 
The roof tile is NOT AT REST when the observer first observes it.

First, it falls from the roof to the top of the window, gaining some velocity. Next, it falls past the window, which the observer times at 0.2 seconds.

Its velocity, when the observer first sees it, is NOT zero. You need to solve for this velocity first. You can then use it to solve for the height of the roof above the window.

- Warren
 
Heres the caluclations

s = 1.6 t= .2 a = g = 9.81 v =?
s = vt +at^2
and v turns out to be 6.038
well this is the velocity when it reached the point -lets say- x which is when observer first sees it. because it fell off the roof and traveled let's say distance d already...so...
d = (v^.5)/2g
= 1.86 m

i think that's pretty much it.
 
s = (1/2)gt^2

and

s+1.6 =(1/2)g(t+0.20)^2

think! and eliminate t.

M.P.
 
Not going to give an answer here but the working I would use seems to differ slightly from others, and arrives at a slightly different expression.

we know that distance = 1/2(inital velocity + final velocity) x time or...
s =1/2(u+v)t
we also know that velocity (v) = u + at
using substitution you can end up with an equation that involves only one variable...that is 'u'
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 36 ·
2
Replies
36
Views
3K
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
3
Views
10K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
2K