Easy Vertical Motion Question

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Homework Help Overview

The discussion revolves around a vertical motion problem involving an object thrown from a window at a speed of 18 m/s, with a focus on the time it takes to hit the ground, which is stated to be 1.6 seconds. Participants are questioning the feasibility of the scenario presented in the problem.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of the object being thrown upwards versus downwards and the time taken for each scenario. There are questions about the validity of the time frame given the initial velocity and the effects of gravity.

Discussion Status

The discussion is active, with various interpretations being explored. Some participants provide calculations based on different assumptions about the direction of the throw, while others express skepticism about the problem's setup. A participant notes that the question was removed from the test after discussing it with the teacher.

Contextual Notes

There is a lack of clarity regarding whether the object was thrown upwards or downwards, which significantly affects the analysis. The problem also raises questions about the physical constraints of the scenario, such as the height of the window relative to the ground.

Inquiring_Mike
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"Easy" Vertical Motion Question

Hey,
This question was on my test today, but it didnt seem right to me..
Q) An object is thrown vertically at 18m/s from a window and hits the ground 1.6s later. What is the height of the window above the ground?


Is this even possible?
 
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Vertically up or down?

Use s = ut + 0.5 a*t^2

You got the t, you got the u, you got the a.

Seems pretty possible... :wink:
 
Vertically up.
But how can the entire trip only take 1.6s if the object is shot upwards at 18m/s? Wouldn't it take at least that long to slow the ball down so it could work on its down trip?
 
Were you specifically told "up"? If an object is thrown upwards at 18 m/s, then it will take 18/g= 18/9.8= 1.8 seconds for the object to reach its highest point. Obviously it can't hit the ground before that.

On the other hand, if the object is thrown vertically downwards at 18 m/s, the distance traveled at time t is x= 18t+ (g/2)t2.
With t= 1.6 seconds, g= 9.8 m/s2, that gives
x= 18(1.6)+ 4.9(1.6)2= 41.3 meters.

Another possibility is that the window is below ground!
Throwing an object up at 18 m/s gives a height of x= 18t- (g/2)t2. With t= 1.6 and g= 9.8, that gives x= 18(1.6)- (4.9)(1.6)2= 16.3 meters. The ground is 16.3 meters above the window!
 
Thanks for the help everyone!
It was up and I told my teacher and he took the question off the test...
 

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